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11.3E: Exercises for The Dot Product - Mathematics


For exercises 1-4, the vectors (vecs{u}) and (vecs{v}) are given. Calculate the dot product (vecs{u}cdotvecs{v}).

1) (quad vecs{u}=⟨3,0⟩, quad vecs{v}=⟨2,2⟩)

Answer:
(vecs{u}cdotvecs{v}=6)

2) (quad vecs{u}=⟨3,−4⟩, quad vecs{v}=⟨4,3⟩)

3) (quad vecs{u}=⟨2,2,−1⟩, quad vecs{v}=⟨−1,2,2⟩)

Answer:
(vecs{u}cdotvecs{v}=0)

4) (quad vecs{u}=⟨4,5,−6⟩, quad vecs{v}=⟨0,−2,−3⟩)

For exercises 5-8, the vectors (vecs{a}, ,vecs{b}), and (vecs{c}) are given. Determine the vectors ((vecs{a}cdotvecs{b})vecs{c}) and ((vecs{a}⋅vecs{c})vecs{b}.) Express the vectors in component form.

5) (quad vecs{a}=⟨2,0,−3⟩, quad vecs{b}=⟨−4,−7,1⟩, quad vecs{c}=⟨1,1,−1⟩)

Answer:
( (vecs a⋅vecs b)vecs c=⟨−11,−11,11⟩; (vecs a⋅vecs c)b=⟨−20,−35,5⟩)

6) (quad vecs{a}=⟨0,1,2⟩, quad vecs{b}=⟨−1,0,1⟩, quad vecs{c}=⟨1,0,−1⟩)

7) (quad vecs{a}=mathbf{hat i} +mathbf{hat j} , quad vecs{b}=mathbf{hat i} −mathbf{hat k} , quad vecs{c}=mathbf{hat i} −2mathbf{hat k} )

Answer:
( (vecs a⋅vecs b)vecs c=⟨1,0,−2⟩; (vecs a⋅vecs c)vecs b=⟨1,0,−1⟩)

8) (quad vecs{a}=mathbf{hat i} −mathbf{hat j} +mathbf{hat k} , quad vecs{b}=mathbf{hat j} +3mathbf{hat k} , quad vecs{c}=−mathbf{hat i} +2mathbf{hat j} −4mathbf{hat k} )

For exercises 9-12, two vectors are given.

a. Find the measure of the angle (θ) between these two vectors. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

b. Is (θ) an acute angle?

9) [T] (quad vecs{a}=⟨3,−1⟩, quad vecs{b}=⟨−4,0⟩)

Answer:
( a. θ=2.82)rad; ( b. θ) is not acute.

10) [T] (quad vecs{a}=⟨2,1⟩, quad vecs{b}=⟨−1,3⟩)

11) (quad vecs{u}=3i, quad vecs{v}=4mathbf{hat i} +4mathbf{hat j} )

Answer:
( a. θ=frac{π}{4})rad; ( b. θ) is acute.

12) (quad vecs{u}=5i, quad vecs{v}=−6mathbf{hat i} +6mathbf{hat j} )

For exercises 13-18, find the measure of the angle between the three-dimensional vectors (vecs{a}) and (vecs{b}). Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

13) (quad vecs{a}=⟨3,−1,2⟩, quad vecs{b}=⟨1,−1,−2⟩)

Answer:
( θ=frac{π}{2})

14) (quad vecs{a}=⟨0,−1,−3⟩, quad vecs{b}=⟨2,3,−1⟩)

15) (quad vecs{a}=mathbf{hat i} +mathbf{hat j} , quad vecs{b}=mathbf{hat j} −mathbf{hat k} )

Answer:
( θ=frac{π}{3})

16) (quad vecs{a}=mathbf{hat i} −2mathbf{hat j} +mathbf{hat k} , quad vecs{b}=mathbf{hat i} +mathbf{hat j} −2mathbf{hat k} )

17) [T] (quad vecs{a}=3mathbf{hat i} −mathbf{hat j} −2mathbf{hat k} , quad vecs{b}=vec v+vec w,) where (quad vecs{v}=−2mathbf{hat i} −3mathbf{hat j} +2mathbf{hat k} ) and (vecs{w}=mathbf{hat i} +2mathbf{hat k} )

Answer:
( θ=2) rad

18) [T] (quad vecs{a}=3mathbf{hat i} −mathbf{hat j} +2mathbf{hat k} , quad vecs{b}=vec v−vec w,) where (quad vecs{v}=2mathbf{hat i} +mathbf{hat j} +4mathbf{hat k} ) and (vecs{w}=6mathbf{hat i} +mathbf{hat j} +2mathbf{hat k} )

For exercises 19-22, determine whether the given vectors are orthogonal.

19) (quad vecs{a}=⟨x,y⟩, quad vecs{b}=⟨−y, x⟩), where (x) and (y) are nonzero real numbers

Answer:
Orthogonal

20) (quad vecs{a}=⟨x, x⟩, quad vecs{b}=⟨−y, y⟩), where (x) and (y) are nonzero real numbers

21) (quad vecs{a}=3mathbf{hat i} −mathbf{hat j} −2mathbf{hat k} , quad vecs{b}=−2mathbf{hat i} −3mathbf{hat j} +mathbf{hat k} )

Answer:
Not orthogonal

22) (quad vecs{a}=mathbf{hat i} −mathbf{hat j} , quad vecs{b}=7mathbf{hat i} +2mathbf{hat j} −mathbf{hat k} )

23) Find all two-dimensional vectors a orthogonal to vector ( vecs b=⟨3,4⟩.) Express the answer in component form.

Answer:
( vecs a=⟨−frac{4α}{3},α⟩,) where ( α≠0) is a real number

24) Find all two-dimensional vectors ( vecs a) orthogonal to vector ( vecs b=⟨5,−6⟩.) Express the answer by using standard unit vectors.

25) Determine all three-dimensional vectors ( vecs u) orthogonal to vector ( vecs v=⟨1,1,0⟩.) Express the answer by using standard unit vectors.

Answer:
( vecs u=−αmathbf{hat i} +αmathbf{hat j} +βmathbf{hat k} ,) where ( α) and ( β) are real numbers such that ( α^2+β^2≠0)

26) Determine all three-dimensional vectors (vecs u) orthogonal to vector (vecs v=mathbf{hat i} −mathbf{hat j} −mathbf{hat k} ). Express the answer in component form.

27) Determine the real number (α) such that vectors (vecs a=2mathbf{hat i} +3mathbf{hat j} ) and (vecs b=9mathbf{hat i} +αmathbf{hat j} ) are orthogonal.

Answer:
(α=−6)

28) Determine the real number (α) such that vectors (vecs a=−3mathbf{hat i} +2mathbf{hat j} ) and (vecs b=2mathbf{hat i} +αmathbf{hat j} ) are orthogonal.

29) [T] Consider the points (P(4,5)) and (Q(5,−7)), and note that (O) represents the origin.

a. Determine vectors (vecd{OP}) and (vecd{OQ}). Express the answer by using standard unit vectors.

b. Determine the measure of angle (O) in triangle (OPQ). Express the answer in degrees rounded to two decimal places.

Answer:
(a. vecd{OP}=4mathbf{hat i} +5mathbf{hat j} , quad vecd{OQ}=5mathbf{hat i} −7mathbf{hat j} ; quad b. 105.8°)

30) [T] Consider points ( A(1,1), B(2,−7),) and (C(6,3)).

a. Determine vectors ( vecd{BA}) and (vecd{BC}). Express the answer in component form.

b. Determine the measure of angle (B) in triangle (ABC). Express the answer in degrees rounded to two decimal places.

31) Determine the measure of angle (A) in triangle (ABC), where (A(1,1,8), B(4,−3,−4),) and (C(−3,1,5).) Express your answer in degrees rounded to two decimal places.

Answer:
(68.33°)

32) Consider points (P(3,7,−2)) and (Q(1,1,−3).) Determine the angle between vectors (vecd{OP}) and (vecd{OQ}). [Remember that (O) represents the origin.] Express the answer in degrees rounded to two decimal places.

For exercises 33-34, determine which (if any) pairs of the following vectors are orthogonal.

33) (quadvecs u=⟨3,7,−2⟩, quad vecs v=⟨5,−3,−3⟩, quad vecs w=⟨0,1,−1⟩)

Answer:
(vecs u) and (vecs v) are orthogonal; (vecs v) and (vecs w) are orthogonal, but (vecs u) and (vecs w) are not orthogonal.

34) (quadvecs u=mathbf{hat i} −mathbf{hat k} , quad vecs v=5mathbf{hat j} −5mathbf{hat k} , quad vecs w=10mathbf{hat j} )

35) Use vectors to show that a parallelogram with equal diagonals is a rectangle.

36) Use vectors to show that the diagonals of a rhombus are perpendicular.

37) Show that (vecs u⋅(vecs v+vecs w)=vecs u⋅vecs v+vecs u⋅vecs w) is true for any vectors (vecs u, vecs v), and (vecs w).

38) Verify the identity (vecs u⋅(vecs v+vecs w)=vecs u⋅vecs v+vecs u⋅vecs w) for vectors (vecs u=⟨1,0,4⟩, vecs v=⟨−2,3,5⟩,) and (vecs w=⟨4,−2,6⟩).

For exercises 39-41, determine (vecs ucdotvecs v) using the given information.

39) (quad|vecs u| = 5), (quad|vecs v| = 3), and the angle between (vecs u) and (vecs v) is (pi/6) rad.

Answer:
(vecs u cdot vecs v = (5)(3)cos frac{pi}{6} = 15cdotfrac{sqrt{3}}{2} = frac{15sqrt{3}}{2} )

40) (quad|vecs u| = 20), (quad|vecs v| = 15), and the angle between (vecs u) and (vecs v) is (5pi/4) rad.

41) (quad|vecs u| = 7), (quad|vecs v| = 12), and the angle between (vecs u) and (vecs v) is (pi/2) rad.

Answer:
(vecs u cdot vecs v = (7)(12)cos frac{pi}{2} = 84cdot0 = 0 )

42) Considering the definition of the dot product, (vecs ucdotvecs v = |vecs u||vecs v|cos heta), where ( heta) is the angle between (vecs u) and (vecs v), what can you say about the angle ( heta) between two nonzero vectors (vecs u) and (vecs v) if:

a. (quadvecs u cdot vecs v > 0)?

b. (quadvecs u cdot vecs v < 0)?

c. (quadvecs u cdot vecs v = 0)?

In exercises 43-45, you are given the vertices of a triangle. Use dot products to determine whether each triangle is acute, obtuse, or right.

43) (quad (2, 3, 0), , (3, 1, -2), , (-1, 4, 5))

Answer:
Label the points (P), (Q) and (R). Then determine the vectors (vecd{PQ}) and (vecd{PR}).
Here we see that, if (P = (2, 3, 0), , Q = (3, 1, -2)) and (R = (-1, 4, 5)), so we have (vecd{PQ} = <1,-2,-2>) and (vecd{PR} = <-3,1,5>).
Then (vecd{PQ} cdot vecd{PR} = -15 < 0), so angle (QPR) is obtuse. Therefore this triangle must be an obtuse triangle.

44) (quad (5, 1, 0), , (7, 1, 1), , (6, 3, 2))

45) (quad (6, 1, 4), , (3, 2, -1), , (2, 3, 1))

Solution:
Label the points (P), (Q) and (R). Then determine the vectors (vecd{PQ}) and (vecd{PR}).
Here we see that, if (P = (6, 1, 4), , Q = (3, 2, -1)) and (R = (2, 3, 1)), so we have (vecd{PQ} = <-3,1,-5>) and (vecd{PR} = <-4,2,-3>).
Then (vecd{PQ} cdot vecd{PR} = 29 > 0), so angle (QPR) is acute. This triangle could still be any of the three types, so we need to check another angle.
Considering angle (PQR) we find (vecd{QP} = <3,-1,5>) and (vecd{QR} = <-1,1,2>), so (vecd{QP} cdot vecd{QR} = 6 > 0), showing angle (PQR) is also acute.
We then must check the third angle to determine whether this is a right triangle or an acute triangle.
To check angle (PRQ) we find (vecd{RP} = <4,-2,3>) and (vecd{RQ} = <1,-1,-2>), giving us (vecd{RP} cdot vecd{RQ} = 0). Angle (PRQ) is thus a right angle.
Therefore this triangle is a right triangle.

46) Which of the following operations are allowed for nonzero vectors (vecs u), (vecs v), and (vecs w), and which are not. Explain your reasoning.

a. (vecs u + left(vecs v cdot vecs w ight)) b. (left(vecs u cdot vecs v ight) cdot vecs w) c. (left(vecs u cdot vecs v ight) vecs w)
d. (left(vecs u + vecs v ight) cdot vecs w) e. (left(vecs u + vecs v ight) cdot |vecs w|) f. (|vecs u + vecs v| |vecs w|)

Projections

For the exercises 47-50, given the vectors (vecs u) and (vecs v):

a. Find the vector projection ( ext{Proj}_vecs{u}vecs v) of vector (vecs v) onto vector (vecs u) and the component of (vecs v) that is orthogonal to (vecs u), i.e., (vecs v_ ext{perp}). Express your answers in component form.

b. Find the scalar projection ( ext{comp}_vecs{u}vecs v = | ext{Proj}_vecs{u}vecs v |) of vector (vecs v) onto vector (vecs u).

c. Find the vector projection ( ext{Proj}_vecs{v}vecs u) of vector (vecs u) onto vector (vecs v) and the component of (vecs u) that is orthogonal to (vecs v), i.e., (vecs u_ ext{perp}). Express your answers in unit vector form.

d. Find the scalar projection ( ext{comp}_vecs{v}vecs u) of vector (vecs u) onto vector (vecs v).

47) (quadvecs u=5mathbf{hat i} +2mathbf{hat j} , quad vecs v=2mathbf{hat i} +3mathbf{hat j} )

Answer:
a. ( ext{Proj}_vecs{u}vecs v =⟨frac{80}{29},frac{32}{29}⟩) and (vecs v_ ext{perp} = <-frac{22}{29}, frac{55}{29}>); b. ( ext{comp}_vecs{u}vecs v=frac{16}{sqrt{29}} = frac{16sqrt{29}}{29};)
c. ( ext{Proj}_vecs{v}vecs u = frac{32}{13}mathbf{hat i} + frac{48}{13}mathbf{hat j} ) and (vecs u_ ext{perp} = frac{33}{13}mathbf{hat i} - frac{22}{13}mathbf{hat j} ); d. ( ext{comp}_vecs{v}vecs u=frac{16}{sqrt{13}}=frac{16sqrt{13}}{13})

48) (quadvecs u=⟨−4,7⟩,quad vecs v=⟨3,5⟩)

49) (quadvecs u=3mathbf{hat i} +2mathbf{hat k} , quad vecs v=2mathbf{hat j} +4mathbf{hat k} )

Answer:
a. ( ext{Proj}_vecs{u}vecs v =⟨frac{24}{13},0,frac{16}{13}⟩) and (vecs v_ ext{perp} = <-frac{24}{13}, frac{26}{13}, frac{36}{13}>); b. ( ext{comp}_vecs{u}vecs v=frac{8}{sqrt{13}}=frac{8sqrt{13}}{13})
c. ( ext{Proj}_vecs{v}vecs u =frac{4}{5}mathbf{hat j} + frac{8}{5}mathbf{hat k} ) and (vecs u_ ext{perp} = 3mathbf{hat i} - frac{4}{5}mathbf{hat j} + frac{2}{5}mathbf{hat k} ); d. ( ext{comp}_vecs{v}vecs u=frac{sqrt{80}}{5} = frac{4sqrt{5}}{5};)

50) (quadvecs u=⟨4,4,0⟩, quad vecs v=⟨0,4,1⟩)

51) Consider the vectors (vecs u=4mathbf{hat i} −3mathbf{hat j} ) and (vecs v=3mathbf{hat i} +2mathbf{hat j} .)

a. Find the component form of vector ( ext{Proj}_vecs{u}vecs v) that represents the projection of (vecs v) onto (vecs u).

b. Write the decomposition (vecs v=vecs w+vecs q) of vector (vecs v) into the orthogonal components (vecs w) and (vecs q), where (vecs w) is the projection of (vecs v) onto (vecs u) and (vecs q) is the vector component of (vecs v) orthogonal to the direction of (vecs u). That is, ( vecs q = vecs v_ ext{perp}).

Answer:
a. ( ext{Proj}_vecs{u}vecs v=⟨frac{24}{25},−frac{18}{25}⟩); b. (vecs q = vecs v_ ext{perp} =⟨frac{51}{25},frac{68}{25}⟩),
(vecs v =vecs w+vecs q= ext{Proj}_vecs{u}vecs v + vecs v_ ext{perp} =⟨frac{24}{25},−frac{18}{25}⟩+⟨frac{51}{25},frac{68}{25}⟩)
So we have that, (vecs v = ⟨frac{24}{25},−frac{18}{25}⟩+⟨frac{51}{25},frac{68}{25}⟩).

52) Consider vectors (vecs u=2mathbf{hat i} +4mathbf{hat j} ) and (vecs v=4mathbf{hat j} +2mathbf{hat k} .)

a. Find the component form of vector (vecs w= ext{Proj}_vecs{u}vecs v) that represents the projection of (vecs v) onto (vecs u).

b. Write the decomposition (vecs v=vecs w+vecs q) of vector (vecs v) into the orthogonal components (vecs w) and (vecs q), where (vecs w) is the projection of (vecs v) onto (vecs u) and (vecs q) is a vector orthogonal to the direction of (vecs u).

53) A (50,000)-pound truck is parked on a hill with a (5°) slope above the horizontal (in the positive (x)-direction). Considering only the force due to gravity, find

a. the component of the weight of the truck that is pulling the truck down the hill (aligned with the road)

b. the component of the weight of the truck that is perpendicular to the road.

c. the magnitude of the force needed to keep the truck from rolling down the hill

d. the magnitude of the force that is perpendicular to the road.

Solution:
Let (vecs F) be the force due to gravity on the truck (its weight) and let (vecs r) be the unit vector pointing up the hill to the right.

Then
a. the component of the weight of the truck that is pulling the truck down the hill is ( ext{Proj}_vecs{r}vecs F).
Now (vecs F = -50,000mathbf{hat j} quad ext{and}quad vecs r = cos 5°,mathbf{hat i} + sin 5°,mathbf{hat j} ) and (vecs Fcdotvecs r = -50,000sin 5° approx -4357.78714)
Note that (|vecs r| = 1).
Then: (quad, ext{Proj}_vecs{r}vecs F = left(frac{vecs Fcdotvecs r}{|vecs r|^2} ight)vecs r quad = left(-50,000sin 5° ight)vecs r quad=quad -50000sin 5° cos 5° ,mathbf{hat i} - 50000sin^2 5° ,mathbf{hat j} ).
Using the identity, (sin 2 heta = 2cos heta sin heta), we can simplify this as
( ext{Proj}_vecs{r}vecs F = -25000sin 10° ,mathbf{hat i} - 50000sin^2 5°, mathbf{hat j} quadapproxquad -4341.2,mathbf{hat i} - 379.8,mathbf{hat j} ).

b. the component of the weight of the truck that is perpendicular to the road is (vecs F_ ext{perp}).
Since we know that (vecs F = ext{Proj}_vecs{r}vecs F+vecs F_ ext{perp}), we have that (vecs F_ ext{perp} = vecs F - ext{Proj}_vecs{r}vecs F).
So (quad vecs F_ ext{perp} = -50000mathbf{hat j} - left( -25000sin 10° ,mathbf{hat i} - 50000sin^2 5°, mathbf{hat j} ight))
(= 25000sin 10° ,mathbf{hat i} + left(50000sin^2 5° - 50000 ight)mathbf{hat j} = 25000sin 10° ,mathbf{hat i} - 50000cos^2 5° ,mathbf{hat j} quadapproxquad 4341.2,mathbf{hat i} - 49620.2,mathbf{hat j} ).

c. the magnitude of the force needed to keep the truck from rolling down the hill is (| ext{Proj}_vecs{r}vecs F | = 50,000sin 5°) lbs (approx 4357.8) lbs. Note that this force is actually the negative of the projection vector, but its magnitude is the same.

d. the magnitude of the truck's weight that is perpendicular to the road is ( | vecs F_ ext{perp} | approx 49809.7) lbs.

54) Given the vectors (vecs u) and (vecs v) shown in each diagram below, draw in ( ext{Proj}_vecs{v}vecs u) and ( vecs u_ ext{perp} ).

Work

55) Find the work done by force (vecs F=⟨5,6,−2⟩) (measured in Newtons) that moves a particle from point (P(3,−1,0)) to point (Q(2,3,1)) along a straight line (the distance is measured in meters).

Answer:
(17, ext{N⋅m})

56) [T] A sled is pulled by exerting a force of 100 N on a rope that makes an angle of (25°) with the horizontal. Find the work done in pulling the sled 40 m. (Round the answer to one decimal place.)

57) [T] A father is pulling his son on a sled at an angle of (20°) with the horizontal with a force of 25 lb (see the following image). He pulls the sled in a straight path of 50 ft. How much work was done by the man pulling the sled? (Round the answer to the nearest integer.)

Answer:
1175 ft⋅lb

58) [T] A car is towed using a force of 1600 N. The rope used to pull the car makes an angle of 25° with the horizontal. Find the work done in towing the car 2 km. Express the answer in joules ((1) J(=1) N⋅m) rounded to the nearest integer.

59) [T] A boat sails north aided by a wind blowing in a direction of ( N30°E) with a magnitude of 500 lb. How much work is performed by the wind as the boat moves 100 ft? (Round the answer to two decimal places.)

Answer:
(25000sqrt{3}) ft-lbs (approx 43,301.27) ft-lbs
Solution:
Vector representing the wind: (vecs w = 500cos 60^{circ} mathbf{hat i} + 500sin 60^{circ} mathbf{hat j})
Vector representing the displacement to the north: (vecs d = 100 mathbf{hat j})
Work done by the wind: (W = vecs w cdot vecs d = 25000sqrt{3}) ft-lbs (approx 43,301.27) ft-lbs

Other Applications of the Dot Product

60) [T] Find the vectors that join the center of a clock to the hours 1:00, 2:00, and 3:00. Assume the clock is circular with a radius of 1 unit.

61) A methane molecule has a carbon atom situated at the origin and four hydrogen atoms located at points (P(1,1,−1),Q(1,−1,1),R(−1,1,1),) and (S(−1,−1,−1)) (see figure).

a. Find the distance between the hydrogen atoms located at (P)and (R).

b. Find the angle between vectors (vecd{OS}) and (vecd{OR}) that connect the carbon atom with the hydrogen atoms located at (S) and (R), which is also called the bond angle. Express the answer in degrees rounded to two decimal places.

Answer:
(a. 2sqrt{2}; quad b. 109.47°)

62) Vector (vecs p=⟨150,225,375⟩) represents the price of certain models of bicycles sold by a bicycle shop. Vector (vecs n=⟨10,7,9⟩) represents the number of bicycles sold of each model, respectively. Compute the dot product (vecs p⋅vecs n) and state its meaning.

63) [T] Two forces (vecs F_1) and (vecs F_2) are represented by vectors with initial points that are at the origin. The first force has a magnitude of 20 lb and passes through the point (P(1,1,0)). The second force has a magnitude of 40 lb and passes through the point (Q(0,1,1)). Let (vecs F) be the resultant force of forces (vecs F_1) and (vecs F_2).

a. Find the magnitude of (vecs F). (Round the answer to one decimal place.)

b. Find the direction angles of (vecs F). (Express the answer in degrees rounded to one decimal place.)

Answer:
( a. ∥vecs F_1+vecs F_2∥=52.9) lb; b. The direction angles are (α=74.5°,,β=36.7°,) and ( γ=57.7°.)

64) [T] Consider (vecs r(t)=⟨cos t,sin t,2t⟩) the position vector of a particle at time (t∈[0,30]), where the components of (vecs r) are expressed in centimeters and time in seconds. Let (vecd{OP}) be the position vector of the particle after 1 sec.

a. Show that all vectors (vecd{PQ}), where (Q(x,y,z)) is an arbitrary point, orthogonal to the instantaneous velocity vector (vecs v(1)) of the particle after 1 sec, can be expressed as ( vecd{PQ}=⟨x−cos 1,y−sin 1,z−2⟩), where (xsin 1−ycos 1−2z+4=0.) The set of point (Q) describes a plane called the normal plane to the path of the particle at point (P).

b. Use a CAS to visualize the instantaneous velocity vector and the normal plane at point (P) along with the path of the particle.

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

Exercises and LaTeX edited by Paul Seeburger. Problems 39-46 and 53 & 54 were created by Paul Seeburger.
Solution to Problem 59 also was added by Paul Seeburger.


Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

The goal of this section is to answer the following question. Given two vectors, what is the angle between them?

Since vectors have no position, we are free to place vectors wherever we like. If the two vectors are placed tail-to-tail, there is now a reasonable interpretation of the question: we seek the measure of the smallest angle between the two vectors, in the plane in which they lie. Figure 6.5 illustrates the situation.

Since the angle ( heta) lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines. Remember that the law of cosines states (c^2 = a^2 + b^2 - 2ab cos C ext<.>)

The lengths of the sides of the triangle in Figure 6.5 are (|vect| ext<,>) (|vect| ext<,>) and (|vect-vect| ext<.>) Let (vect=langle v_1,v_2,v_3 angle) and (vect=langle w_1,w_2,w_3 angle ext<>) then

A bit of simple arithmetic with the coordinates of (vect) and (vect) allows us to compute the cosine of the angle between them. If necessary we can use the arccosine to get ( heta ext<,>) but in many problems (cos heta) turns out to be all we really need.

The numerator of the fraction that gives us (cos heta) turns up a lot, so we give it a name and more compact notation: we call it the , and write it as

This is the same symbol we use for ordinary multiplication, but there should never be any confusion you can tell from context whether we are “multiplying” vectors or numbers. (We might also use the dot for scalar multiplication: (acdotvect=avect ext<>) again, it is clear what is meant from context.)

Example 6.4 .

Find the angle between the vectors (vect=langle 1,2,1 angle) and (vect=langle 3,1,-5 angle ext<.>)

We know that (cos heta=vectcdotvect/(|vect||vect|)= (1cdot3 + 2cdot1 + 1cdot(-5))/(|vect||vect|)=0 ext<,>) so ( heta=pi/2 ext<,>) that is, the vectors are perpendicular.

Example 6.5 .

Find the angle between the vectors (vect=langle 3,3,0 angle) and (vect=langle 1,0,0 angle ext<.>)

The following are some special cases worth looking at.

Example 6.6 .

(vect) and (vect<0>=langle 0,0,0 angle)

so the angle between (vect) and itself is zero, which of course is correct.

so the angle is (pi ext<,>) that is, the vectors point in opposite directions, as of course we already knew.

which is undefined. On the other hand, note that since (vectcdotvect<0>=0) it looks at first as if (cos heta) will be zero, which as we have seen means that vectors are perpendicular only when we notice that the denominator is also zero do we run into trouble. One way to “fix” this is to adopt the convention that the zero vector (vect<0>) is perpendicular to all vectors then we can say in general that if (vectcdotvect=0 ext<,>) (vect) and (vect) are perpendicular.

Generalizing the examples, note the following useful facts:

If (vect) is parallel or anti-parallel to (vect) then (vectcdotvect/(|vect||vect|)=pm1 ext<,>) and conversely, if (vectcdotvect/(|vect||vect|)=1 ext<,>) (vect) and (vect) are parallel, while if (vectcdotvect/(|vect||vect|)=-1 ext<,>) (vect) and (vect) are anti-parallel. (Vectors are parallel if they point in the same direction, anti-parallel if they point in opposite directions.)

If (vect) is perpendicular to (vect) then (vectcdotvect/(|vect||vect|)=0 ext<,>) and conversely if (vectcdotvect/(|vect||vect|)=0) then (vect) and (vect) are perpendicular.

Given two vectors, it is often useful to find the of one vector onto the other, because this turns out to have important meaning in many circumstances. More precisely, given (vect) and (vect ext<,>) we seek a vector parallel to (vect) but with length determined by (vect) in a natural way, as shown in Figure 6.6. (vect

) is chosen so that the triangle formed by (vect ext<,>) (vect

ext<,>) and (vect-vect

) is a right triangle.

Figure 6.6. (vect

) is the projection of (vect) onto (vect ext<.>)

Using a little trigonometry, we see that

this is sometimes called the . To get (vect

) itself, we multiply this length by a vector of length one parallel to (vect ext<:>)

Be sure that you understand why (vect/|vect|) is a vector of length one (also called a ) parallel to (vect ext<.>)

The discussion so far implicitly assumed that (0le hetalepi/2 ext<.>) If (pi/2lt hetalepi ext<,>) the picture is like Figure 6.7. In this case (vectcdot vect) is negative, so the vector

is anti-parallel to (vect ext<,>) and its length is

In general, the scalar projection of (vect) onto (vect) may be positive or negative. If it is negative, it means that the projection vector is anti-parallel to (vect) and that the length of the projection vector is the absolute value of the scalar projection. Of course, you can also compute the length of the projection vector as usual, by applying the distance formula to the vector.

Figure 6.7. (vect

) is the projection of (vect) onto (vect ext<.>)

Note that the phrase “projection onto (vect)” is a bit misleading if taken literally all that (vect) provides is a direction the length of (vect) has no impact on the final vector. In Figure 6.8, for example, (vect) is shorter than the projection vector, but this is perfectly acceptable.

Figure 6.8. (vect

) is the projection of (vect) onto (vect ext<.>)

Physical force is a vector quantity. It is often necessary to compute the “component” of a force acting in a different direction than the force is being applied.

Example 6.7 . Components of Force Vector.

Suppose a ten pound weight is resting on an inclined plane—a pitched roof, for example. Gravity exerts a force of ten pounds on the object, directed straight down. It is useful to think of the component of this force directed down and parallel to the roof, and the component down and directly into the roof. These forces are the projections of the force vector onto vectors parallel and perpendicular to the roof. Suppose the roof is tilted at a (30^circ) angle, as in Figure 6.9. Compute the component of the force directed down the roof and the component of the force directed into the roof.

A vector parallel to the roof is (langle-sqrt3,-1 angle ext<,>) and a vector perpendicular to the roof is (langle 1,-sqrt3 angle ext<.>) The force vector is (vect=langle 0,-10 angle ext<.>) The component of the force directed down the roof is then

with length 5. The component of the force directed into the roof is

with length (5sqrt3 ext<.>) Thus, a force of 5 pounds is pulling the object down the roof, while a force of (5sqrt3) pounds is pulling the object into the roof.

The dot product has some familiar-looking properties that will be useful later, so we list them here. These may be proved by writing the vectors in coordinate form and then performing the indicated calculations subsequently it can be easier to use the properties instead of calculating with coordinates.

Theorem 6.8 . Dot Product Properties.

If (vect ext<,>) (vect ext<,>) and (vect) are vectors and (a) is a real number, then


11.3E: Exercises for The Dot Product - Mathematics

Here's a question whose answer turns out to be very useful: Given two vectors, what is the angle between them?

It may not be immediately clear that the question makes sense, but it's not hard to turn it into a question that does. Since vectors have no position, we are as usual free to place vectors wherever we like. If the two vectors are placed tail-to-tail, there is now a reasonable interpretation of the question: we seek the measure of the smallest angle between the two vectors, in the plane in which they lie. Figure 14.3.1 illustrates the situation.

Since the angle $ heta$ lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines. The lengths of the sides of the triangle in figure 14.3.1 are $|<f A>|$, $|<f B>|$, and $|<f A>-<f B>|$. Let $ds <f A>=langle a_1,a_2,a_3 angle$ and $ds <f B>=langle b_1,b_2,b_3 angle$ then $eqalign< |<f A>-<f B>|^2&=|<f A>|^2+|<f B>|^2-2|<f A>||<f B>|cos hetacr 2|<f A>||<f B>|cos heta&=|<f A>|^2+|<f B>|^2-|<f A>-<f B>|^2cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2-(a_1-b_1)^2-(a_2-b_2)^2-(a_3-b_3)^2cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2cr &qquad-(a_1^2-2a_1b_1+b_1^2) -(a_2^2-2a_2b_2+b_2^2)-(a_3^2-2a_3b_3+b_3^2)cr &=2a_1b_1+2a_2b_2+2a_3b_3cr |<f A>||<f B>|cos heta&=a_1b_1+a_2b_2+a_3b_3cr cos heta&=(a_1b_1+a_2b_2+a_3b_3)/(|<f A>||<f B>|)cr >$ So a bit of simple arithmetic with the coordinates of $f A$ and $f B$ allows us to compute the cosine of the angle between them. If necessary we can use the arccosine to get $ heta$, but in many problems $cos heta$ turns out to be all we really need.

The numerator of the fraction that gives us $cos heta$ turns up a lot, so we give it a name and more compact notation: we call it the dot product, and write it as $<f A>cdot <f B>= a_1b_1+a_2b_2+a_3b_3.$ This is the same symbol we use for ordinary multiplication, but there should never be any confusion you can tell from context whether we are "multiplying'' vectors or numbers. (We might also use the dot for scalar multiplication: $acdot<f V>=a<f V>$ again, it is clear what is meant from context.)

Example 14.3.1 Find the angle between the vectors $<f A>=langle 1,2,1 angle$ and $<f B>=langle 3,1,-5 angle$. We know that $cos heta=<f A>cdot<f B>/(|<f A>||<f B>|)= (1cdot3 + 2cdot1 + 1cdot(-5))/(|<f A>||<f B>|)=0$, so $ heta=pi/2$, that is, the vectors are perpendicular.

Example 14.3.2 Find the angle between the vectors $<f A>=langle 3,3,0 angle$ and $<f B>=langle 1,0,0 angle$. We compute $eqalign< cos heta &= (3cdot1 + 3cdot0 + 0cdot0)/(sqrt<9+9+0>sqrt<1+0+0>)cr &= 3/sqrt <18>= 1/sqrt2cr>$ so $ heta=pi/4$.

Example 14.3.3 Some special cases are worth looking at: Find the angles between $<f A>$ and $<f A>$ $<f A>$ and $<f -A>$ $<f A>$ and $<f 0>=langle 0,0,0 angle$.

$ds cos heta= <f A>cdot<f A>/(|<f A>||<f A>|)=(a_1^2+a_2^2+a_3^2)/ (sqrtsqrt)=1$, so the angle between $<f A>$ and itself is zero, which of course is correct.

$dscos heta=<f A>cdot<f -A>/(|<f A>||<f -A>|)=(-a_1^2-a_2^2-a_3^2)/ (sqrtsqrt)=-1$, so the angle is $pi$, that is, the vectors point in opposite directions, as of course we already knew.

$ds cos heta= <f A>cdot<f 0>/(|<f A>||<f 0>|)=(0+0+0)/ (sqrtsqrt<0^2+0^2+0^2>)$, which is undefined. On the other hand, note that since $<f A>cdot<f 0>=0$ it looks at first as if $cos heta$ will be zero, which as we have seen means that vectors are perpendicular only when we notice that the denominator is also zero do we run into trouble. One way to "fix'' this is to adopt the convention that the zero vector $<f 0>$ is perpendicular to all vectors then we can say in general that if $<f A>cdot<f B>=0$, $f A$ and $f B$ are perpendicular.

Generalizing the examples, note the following useful facts:

1. If $f A$ is parallel or anti-parallel to $f B$ then $<f A>cdot<f B>/(|<f A>||<f B>|)=pm1$, and conversely, if $<f A>cdot<f B>/(|<f A>||<f B>|)=1$, $f A$ and $f B$ are parallel, while if $<f A>cdot<f B>/(|<f A>||<f B>|)=-1$, $f A$ and $f B$ are anti-parallel. (Vectors are parallel if they point in the same direction, anti-parallel if they point in opposite directions.)

2. If $f A$ is perpendicular to $f B$ then $<f A>cdot<f B>/(|<f A>||<f B>|)=0$, and conversely if $<f A>cdot<f B>/(|<f A>||<f B>|)=0$ then $f A$ and $f B$ are perpendicular.

Given two vectors, it is often useful to find the projection of one vector onto the other, because this turns out to have important meaning in many circumstances. More precisely, given $<f A>$ and $<f B>$, we seek a vector parallel to $f B$ but with length determined by $f A$ in a natural way, as shown in figure 14.3.2. $f V$ is chosen so that the triangle formed by $f A$, $f V$, and $<f A>-<f V>$ is a right triangle.

Using a little trigonometry, we see that $ |<f V>|=|<f A>|cos heta= |<f A>|<<f A>cdot<f B>over|<f A>||<f B>|>= <<f A>cdot<f B>over|<f B>|> $ this is sometimes called the scalar projection of $f A$ onto $f B$. To get $f V$ itself, we multiply this length by a vector of length one parallel to $f B$: $ <f V>= <<f A>cdot<f B>over|<f B>|><<f B>over|<f B>|>= <<f A>cdot<f B>over|<f B>|^2><f B>. $ Be sure that you understand why $<f B>/|<f B>|$ is a vector of length one (also called a unit vector) parallel to $f B$.

The discussion so far implicitly assumed that le hetalepi/2$. If $pi/2 Figure 14.3.3. $f V$ is the projection of $f A$ onto $f B$.

Note that the phrase "projection onto $f B

APEX Calculus

The previous section introduced vectors and described how to add them together and how to multiply them by scalars. This section introduces a multiplication on vectors called the dot product.

Definition 10.3.1 . Dot Product.

Let (vec u = la u_1,u_2 a) and (vec v = la v_1,v_2 a) in (mathbb^2 ext<.>) The dot product of (vec u) and (vec v ext<,>) denoted (dotp uv ext<,>) is

Let (vec u = la u_1,u_2,u_3 a) and (vec v = la v_1,v_2,v_3 a) in (mathbb^3 ext<.>) The dot product of (vec u) and (vec v ext<,>) denoted (dotp uv ext<,>) is

Note how this product of vectors returns a scalar, not another vector. We practice evaluating a dot product in the following example, then we will discuss why this product is useful.

Example 10.3.2 . Evaluating dot products.

Let (vec u=la 1,2 a ext<,>) (vec v=la 3,-1 a) in (mathbb^2 ext<.>) Find (dotp uv ext<.>)

Let (vec x = la 2,-2,5 a) and (vec y = la -1, 0, 3 a) in (mathbb^3 ext<.>) Find (dotp xy ext<.>)

Using the definition, we have

The dot product, as shown by the preceding example, is very simple to evaluate. It is only the sum of products. While the definition gives no hint as to why we would care about this operation, there is an amazing connection between the dot product and angles formed by the vectors. Before stating this connection, we give a theorem stating some of the properties of the dot product.

Theorem 10.3.3 . Properties of the Dot Product.

Let (vec u ext<,>) (vec v) and (vec w) be vectors in (mathbb^2) or (mathbb^3) and let (c) be a scalar.

(vec ucdot(vec v+vec w) = dotp uv + dotp uw)

(displaystyle c(dotp uv) = (cvec u)cdot vec v = vec u cdot (cvec v))

(displaystyle dotp vv= orm^2)

The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. Our definition and theorem give properties of the dot product, but we are still likely wondering “What does the dot product mean?” It is helpful to understand that the dot product of a vector with itself is connected to its magnitude.

The next theorem extends this understanding by connecting the dot product to magnitudes and angles. Given vectors (vec u) and (vec v) in the plane, an angle ( heta) is clearly formed when (vec u) and (vec v) are drawn with the same initial point as illustrated in Figure 10.3.4.(a). (We always take ( heta) to be the angle in ([0,pi]) as two angles are actually created.)

The same is also true of 2 vectors in space: given (vec u) and (vec v) in (mathbb^3) with the same initial point, there is a plane that contains both (vec u) and (vec v ext<.>) (When (vec u) and (vec v) are co-linear, there are infinitely many planes that contain both vectors.) In that plane, we can again find an angle ( heta) between them (and again, (0leq hetaleq pi)). This is illustrated in Figure 10.3.4.(b).

The following theorem connects this angle ( heta) to the dot product of (vec u) and (vec v ext<.>)

Theorem 10.3.5 . The Dot Product and Angles.

Let (vec u) and (vec v) be nonzero vectors in (mathbb^2) or (mathbb^3 ext<.>) Then

where ( heta ext<,>) (0leq hetaleq pi ext<,>) is the angle between (vec u) and (vec v ext<.>)

Using Theorem 10.3.3, we can rewrite this theorem as

Note how on the left hand side of the equation, we are computing the dot product of two unit vectors. Recalling that unit vectors essentially only provide direction information, we can informally restate Theorem 10.3.5 as saying “The dot product of two directions gives the cosine of the angle between them.”

When ( heta) is an acute angle (i.e., (0leq heta lt pi/2)), (cos( heta)) is positive when ( heta = pi/2 ext<,>) (cos( heta) = 0 ext<>) when ( heta) is an obtuse angle ((pi/2lt heta leq pi)), (cos( heta)) is negative. Thus the sign of the dot product gives a general indication of the angle between the vectors, illustrated in Figure 10.3.6.

We can use Theorem 10.3.5 to compute the dot product, but generally this theorem is used to find the angle between known vectors (since the dot product is generally easy to compute). To this end, we rewrite the theorem's equation as

We practice using this theorem in the following example.

Example 10.3.7 . Using the dot product to find angles.

Let (vec u = la 3,1 a ext<,>) (vec v = la -2,6 a) and (vec w = la -4,3 a ext<,>) as shown in Figure 10.3.8. Find the angles (alpha ext<,>) (eta) and ( heta ext<.>)

We start by computing the magnitude of each vector.

We now apply Theorem 10.3.5 to find the angles.

We see from our computation that (alpha + eta = heta ext<,>) as indicated by Figure 10.3.8. While we knew this should be the case, it is nice to see that this non-intuitive formula indeed returns the results we expected.

We do a similar example next in the context of vectors in space.

Example 10.3.9 . Using the dot product to find angles.

Let (vec u = la 1,1,1 a ext<,>) (vec v = la -1,3,-2 a) and (vec w = la -5,1,4 a ext<,>) as illustrated in Figure 10.3.10. Find the angle between each pair of vectors.

Between (vec u) and (vec v ext<:>)

Between (vec u) and (vec w ext<:>)

Between (vec v) and (vec w ext<:>)

While our work shows that each angle is (pi/2 ext<,>) i.e., (90^circ ext<,>) none of these angles looks to be a right angle in Figure 10.3.10. Such is the case when drawing three-dimensional objects on the page.

All three angles between these vectors was (pi/2 ext<,>) or (90^circ ext<.>) We know from geometry and everyday life that (90^circ) angles are “nice” for a variety of reasons, so it should seem significant that these angles are all (pi/2 ext<.>) Notice the common feature in each calculation (and also the calculation of (alpha) in Example 10.3.7): the dot products of each pair of angles was 0. We use this as a basis for a definition of the term orthogonal, which is essentially synonymous to perpendicular.

Definition 10.3.11 . Orthogonal.

Nonzero vectors (vec u) and (vec v) are if their dot product is 0.

The term perpendicular originally referred to lines. As mathematics progressed, the concept of “being at right angles to” was applied to other objects, such as vectors and planes, and the term orthogonal was introduced. It is especially used when discussing objects that are hard, or impossible, to visualize: two vectors in 5-dimensional space are orthogonal if their dot product is 0. It is not wrong to say they are perpendicular, but common convention gives preference to the word orthogonal.

Example 10.3.12 . Finding orthogonal vectors.

Let (vec u = la 3,5 a) and (vec v = la 1,2,3 a ext<.>)

Find two vectors in (mathbb^2) that are orthogonal to (vec u ext<.>)

Find two non-parallel vectors in (mathbb^3) that are orthogonal to (vec v ext<.>)

Recall that a line perpendicular to a line with slope (m) has slope (-1/m ext<,>) the “opposite reciprocal slope.” We can think of the slope of (vec u) as (5/3 ext<,>) its “rise over run.” A vector orthogonal to (vec u) will have slope (-3/5 ext<.>) There are many such choices, though all parallel:

There are infinitely many directions in space orthogonal to any given direction, so there are an infinite number of non-parallel vectors orthogonal to (vec v ext<.>) Since there are so many, we have great leeway in finding some. One way is to arbitrarily pick values for the first two components, leaving the third unknown. For instance, let (vec v_1 = la 2,7,z a ext<.>) If (vec v_1) is to be orthogonal to (vec v ext<,>) then (vec v_1cdotvec v = 0 ext<,>) so

So (vec v_1 = la 2, 7, -16/3 a) is orthogonal to (vec v ext<.>) We can apply a similar technique by leaving the first or second component unknown. Another method of finding a vector orthogonal to (vec v) mirrors what we did in part 1. Let (vec v_2 = la-2,1,0 a ext<.>) Here we switched the first two components of (vec v ext<,>) changing the sign of one of them (similar to the “opposite reciprocal” concept before). Letting the third component be 0 effectively ignores the third component of (vec v ext<,>) and it is easy to see that

Clearly (vec v_1) and (vec v_2) are not parallel.

An important construction is illustrated in Figure 10.3.13, where vectors (vec u) and (vec v) are sketched. In Figure 10.3.13.(a), a dotted line is drawn from the tip of (vec u) to the line containing (vec v ext<,>) where the dotted line is orthogonal to (vec v ext<.>) In Figure 10.3.13.(b), the dotted line is replaced with the vector (vec z) and (vec w) is formed, parallel to (vec v ext<.>) It is clear by the diagram that (vec u = vec w+vec z ext<.>) What is important about this construction is this: (vec u) is decomposed as the sum of two vectors, one of which is parallel to (vec v) and one that is perpendicular to (vec v ext<.>) It is hard to overstate the importance of this construction (as we'll see in upcoming examples).

The vectors (vec w ext<,>) (vec z) and (vec u) as shown in Figure 10.3.13.(b) form a right triangle, where the angle between (vec v) and (vec u) is labeled ( heta ext<.>) We can find (vec w) in terms of (vec v) and (vec u ext<.>)

Using trigonometry, we can state that

We also know that (vec w) is parallel to to (vec v) that is, the direction of (vec w) is the direction of (vec v ext<,>) described by the unit vector (vec v/ orm ext<.>) The vector (vec w) is the vector in the direction (vec v/ orm) with magnitude ( ormcos( heta) ext<:>)

Since this construction is so important, it is given a special name.

Definition 10.3.14 . Orthogonal Projection.

Let nonzero vectors (vec u) and (vec v) be given. The of (vec u) onto (vec v ext<,>) denoted (proj uv ext<,>) is

Example 10.3.15 . Computing the orthogonal projection.

Let (vec u= la -2,1 a) and (vec v=la 3,1 a ext<.>) Find (proj uv ext<,>) and sketch all three vectors with initial points at the origin.

Let (vec w = la 2,1,3 a) and (vec x = la 1,1,1 a ext<.>) Find (proj wx ext<,>) and sketch all three vectors with initial points at the origin.

Vectors (vec u ext<,>) (vec v) and (proj uv) are sketched in Figure 10.3.16. Note how the projection is parallel to (vec v ext<>) that is, it lies on the same line through the origin as (vec v ext<,>) although it points in the opposite direction. That is because the angle between (vec u) and (vec v) is obtuse (i.e., greater than (90^circ)).

These vectors are sketched in Figure 10.3.17.(a), and again in Figure 10.3.17.(b) from a different perspective. Because of the nature of graphing these vectors, the sketch in Figure 10.3.17.(a) makes it difficult to recognize that the drawn projection has the geometric properties it should. The graph shown in Figure 10.3.17.(b) illustrates these properties better.

We can use the properties of the dot product found in Theorem 10.3.3 to rearrange the formula found in Definition 10.3.14:

The above formula shows that the orthogonal projection of (vec u) onto (vec v) is only concerned with the direction of (vec v ext<,>) as both instances of (vec v) in the formula come in the form (vec v/ orm ext<,>) the unit vector in the direction of (vec v ext<.>)

A special case of orthogonal projection occurs when (vec v) is a unit vector. In this situation, the formula for the orthogonal projection of a vector (vec u) onto (vec v) reduces to just (proj uv = (vec ucdotvec v)vec v ext<,>) as (vec vcdotvec v = 1 ext<.>)

This gives us a new understanding of the dot product. When (vec v) is a unit vector, essentially providing only direction information, the dot product of (vec u) and (vec v) gives “how much of (vec u) is in the direction of (vec v ext<.>)” This use of the dot product will be very useful in future sections.

Now consider Figure 10.3.18 where the concept of the orthogonal projection is again illustrated. It is clear that

As we know what (vec u) and (proj uv) are, we can solve for (vec z) and state that

This leads us to rewrite Equation (10.3.2) in a seemingly silly way:

This is not nonsense, as pointed out in the following Key Idea. (Notation note: the expression “(parallel vec y)” means “is parallel to (vec y ext<.>)” We can use this notation to state “(vec xparallelvec y)” which means “(vec x) is parallel to (vec y ext<.>)” The expression “(perp vec y)” means “is orthogonal to (vec y ext<,>)” and is used similarly.)

Key Idea 10.3.19 . Orthogonal Decomposition of Vectors.

Let nonzero vectors (vec u) and (vec v) be given. Then (vec u) can be written as the sum of two vectors, one of which is parallel to (vec v ext<,>) and one of which is orthogonal to (vec v ext<:>)

We illustrate the use of this equality in the following example.

Example 10.3.20 . Orthogonal decomposition of vectors.

Let (vec u = la -2,1 a) and (vec v = la 3,1 a) as in Example 10.3.15. Decompose (vec u) as the sum of a vector parallel to (vec v) and a vector orthogonal to (vec v ext<.>)

Let (vec w =la 2,1,3 a) and (vec x =la 1,1,1 a) as in Example 10.3.15. Decompose (vec w) as the sum of a vector parallel to (vec x) and a vector orthogonal to (vec x ext<.>)

In Example 10.3.15, we found that (proj uv = la -1.5,-0.5 a ext<.>) Let

Is (vec z) orthogonal to (vec v) ? (i.e., is (vec z perpvec v) ?) We check for orthogonality with the dot product:

Since the dot product is 0, we know (vec z perp vec v ext<.>) Thus:

We found in Example 10.3.15 that (proj wx = la 2,2,2 a ext<.>) Applying the Key Idea, we have:

We check to see if (vec z perp vec x ext<:>)

Since the dot product is 0, we know the two vectors are orthogonal. We now write (vec w) as the sum of two vectors, one parallel and one orthogonal to (vec x ext<:>)

We give an example of where this decomposition is useful.

Example 10.3.21 . Orthogonally decomposing a force vector.

Consider Figure 10.3.22.(a), showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. Find the components of force, and their magnitudes, acting on the box (as sketched in Figure 10.3.22.(b)):

in the direction of the ramp, and

As the ramp rises 5ft over a horizontal distance of 20ft, we can represent the direction of the ramp with the vector (vec r= la 20,5 a ext<.>) Gravity pulls down with a force of 50lb, which we represent with (vec g = la 0,-50 a ext<.>)

To find the force of gravity in the direction of the ramp, we compute (proj gr ext<:>)

The magnitude of (proj gr) is ( orm = 50/sqrt <17>approx 12.13 ext< lb > ext<.>) Though the box weighs 50lb, a force of about 12lb is enough to keep the box from sliding down the ramp.

To find the component (vec z) of gravity orthogonal to the ramp, we use Key Idea 10.3.19.

The magnitude of this force is ( orm approx 48.51)lb. In physics and engineering, knowing this force is important when computing things like static frictional force. (For instance, we could easily compute if the static frictional force alone was enough to keep the box from sliding down the ramp.)

Subsection 10.3.1 Application to Work

In physics, the application of a force (F) to move an object in a straight line a distance (d) produces work the amount of work (W) is (W=Fd ext<,>) (where (F) is in the direction of travel). The orthogonal projection allows us to compute work when the force is not in the direction of travel.

Consider Figure 10.3.23, where a force (vec F) is being applied to an object moving in the direction of (vec d ext<.>) (The distance the object travels is the magnitude of (vec d ext<.>)) The work done is the amount of force in the direction of (vec d ext<,>) ( orm ext<,>) times (vnorm d ext<:>)

The expression (dotp Fd) will be positive if the angle between (vec F) and (vec d) is acute when the angle is obtuse (hence (dotp Fd) is negative), the force is causing motion in the opposite direction of (vec d ext<,>) resulting in “negative work.” We want to capture this sign, so we drop the absolute value and find that (W = dotp Fd ext<.>)

Definition 10.3.24 . Work.

Let (vec F) be a constant force that moves an object in a straight line from point (P) to point (Q ext<.>) Let (vec d = overrightarrow ext<.>) The (W) done by (vec F) along (vec d) is (W = dotp Fd ext<.>)

Example 10.3.25 . Computing work.

A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in Figure 10.3.26. Compute the work done.

The figure indicates that the force applied makes a (30^circ) angle with the horizontal, so (vec F = 50la cos(30^circ) ,sin(30^circ) a approx la 43.3,25 a ext<.>) The ramp is represented by (vec d = la 15,3 a ext<.>) The work done is simply

Note how we did not actually compute the distance the object traveled, nor the magnitude of the force in the direction of travel this is all inherently computed by the dot product!

The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. The next section explores another “product” on vectors, the cross product. Once again, angles play an important role, though in a much different way.

Exercises 10.3.2 Exercises

The dot product of two vectors is a , not a vector.

How are the concepts of the dot product and vector magnitude related?

How can one quickly tell if the angle between two vectors is acute or obtuse?

Give a synonym for “orthogonal.”

In the following exercises, find the dot product of the given vectors.

(vec u = la 2,-4 a ext<,>) (vec v = la 3,7 a)

(vec u = la 5,3 a ext<,>) (vec v = la 6,1 a)

(vec u = la 1,-1,2 a ext<,>) (vec v = la 2,5,3 a)

(vec u = la 3,5,-1 a ext<,>) (vec v = la 4,-1,7 a)

(vec u = la 1,1 a ext<,>) (vec v = la 1,2,3 a)

(vec u = la 1,2,3 a ext<,>) (vec v = la 0,0,0 a)

Create your own vectors (vec u ext<,>) (vec v) and (vec w) in (mathbb^2) and show that (vec ucdot (vec v+vec w) = vec ucdot vec v + vec ucdot vec w ext<.>)

Create your own vectors (vec u) and (vec v) in (mathbb^3) and scalar (c) and show that (c(vec ucdot vec v) = vec ucdot (cvec v) ext<.>)

In the following exercises, find the measure of the angle between the two vectors in both radians and degrees.

The angle between (vec u = la 1,1 a) and (vec v = la 1,2 a) is .

The angle between (vec u = la -2,1 a) and (vec v = la 3,5 a) is .

The angle between (vec u = la 8,1,-4 a) and (vec v = la 2,2,0 a) is .

The angle between (vec u = la 1,7,2 a) and (vec v = la 4,-2,5 a) is .

In the following exercises, a vector (vec v) is given. Give two vectors that are orthogonal to (vec v ext<.>)

Find two nonzero vectors orthogonal to (vec v = la 4,7 a ext<.>)

Find two nonzero vectors orthogonal to (vec v = la -3,5 a ext<.>)

Find two nonzero vectors orthogonal to (vec v = la 1,1,1 a ext<.>)

Find two nonzero vectors orthogonal to (vec v = la 1,-2,3 a ext<.>)

In the following exercises, vectors (vec u) and (vec v) are given. Find (proj uv ext<,>) the orthogonal projection of (vec u) onto (vec v ext<,>) and sketch all three vectors with the same initial point.

If (vec u = la 1,2 a) and (vec v = la -1,3 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

If (vec u = la 5,5 a) and (vec v = la 1,3 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

If (vec u = la -3,2 a) and (vec v = la 1,1 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

If (vec u = la -3,2 a) and (vec v = la 2,3 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

If (vec u = la 1,5,1 a) and (vec v = la 1,2,3 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

If (vec u = la 3,-1,2 a) and (vec v = la 2,2,1 a ext<,>) then (proj uv=) .

Sketch all three vectors on the same axes.

In the following exercises, vectors (vec u) and (vec v) are given. Write (vec u) as the sum of two vectors, one of which is parallel to (vec v) (or is zero) and one of which is orthogonal to (vec v ext<.>) Note: these are the same pairs of vectors as found in Exercises 10.3.2.21 — Exercise 10.3.2.26.

Write (vec u = la 1,2 a) as the sum of two vectors, one parallel to (vec v = la -1,3 a) (or zero) and the other perpendicular.

Write (vec u = la 5,5 a) as the sum of two vectors, one parallel to (vec v = la 1,3 a) (or zero) and the other perpendicular.

Write (vec u = la -3,2 a) as the sum of two vectors, one parallel to (vec v = la 1,1 a) (or zero) and the other perpendicular.

Write (vec u = la -3,2 a) as the sum of two vectors, one parallel to (vec v = la 2,3 a) (or zero) and the other perpendicular.

Write (vec u = la 1,5,1 a) as the sum of two vectors, one parallel to (vec v = la 1,2,3 a) (or zero) and the other perpendicular.

Write (vec u = la 3,-1,2 a) as the sum of two vectors, one parallel to (vec v = la 2,2,1 a) (or zero) and the other perpendicular.

A 10lb box sits on a ramp that rises 4ft over a distance of 20ft. How much force is required to keep the box from sliding down the ramp?

A 10lb box sits on a 15ft ramp that makes a (30^circ) angle with the horizontal. How much force is required to keep the box from sliding down the ramp?

How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of (45^circ) to the horizontal?

How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of (10^circ) to the horizontal?

How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied horizontally?

How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied at an angle of (45^circ) to the horizontal?

How much work is performed in moving a box up the length of a 10ft ramp that makes a (5^circ) angle with the horizontal, with 50lb of force applied in the direction of the ramp?


11.3E: Exercises for The Dot Product - Mathematics

Given two vectors v &LongRightArrow and w &LongRightArrow whose components are elements of ℝ , with the same number of components, we define their dot product, written as v &LongRightArrow · w &LongRightArrow or ( v &LongRightArrow , w &LongRightArrow ) as the sum of the products of corresponding components: ∑ i v i w i .

Obvious facts: the dot product is linear in v &LongRightArrow and in w &LongRightArrow and is symmetric between them.

We define the length of v &LongRightArrow to be the positive square root of ( v &LongRightArrow , v &LongRightArrow ) the length of v &LongRightArrow is usually denoted by | v &LongRightArrow | .

Wonderful Fact: the dot product is invariant under rotation of coordinates.

Exercises 3.1 Prove this statement. Solution

As a consequence of this fact, in evaluating v &LongRightArrow · w &LongRightArrow , we can rotate coordinates so that the first basis vector is in the direction of v &LongRightArrow and the second one is perpendicular to it in the plane of v &LongRightArrow and w &LongRightArrow .

Then v &LongRightArrow will have first two coordinates ( | v &LongRightArrow | , 0 ) and if the angle between v &LongRightArrow and w &LongRightArrow is θ , w &LongRightArrow will have ( | w &LongRightArrow | cos ⁡ θ , | w &LongRightArrow | sin ⁡ θ ) as its similarly defined coordinates.
The dot product v &LongRightArrow · w &LongRightArrow
therefore is | v &LongRightArrow | | w &LongRightArrow | cos ⁡ θ , in this coordinate system (that is, with these basis vectors), and hence in any coordinate system obtained by rotations from it.

The fact that the dot product is linear in each of its arguments is extremely important and valuable. It means that you can apply the distributive law in either argument to express the dot product of a sum or difference as the sum or difference of the dot products.

Exercises 3.2 Express the square of the area of a parallelogram with sides v &LongRightArrow and w &LongRightArrow in terms of dot products. Solution

The dot product of v &LongRightArrow and w &LongRightArrow divided by the magnitude of w &LongRightArrow , which is | v &LongRightArrow | cos ⁡ θ , is called the component of v &LongRightArrow in the direction of w &LongRightArrow .

The vector in the w &LongRightArrow direction having magnitude and sign of | v &LongRightArrow | cos ⁡ θ is called the projection of v &LongRightArrow on w &LongRightArrow .

The vector obtained by subtracting the projection of v &LongRightArrow on w &LongRightArrow from v &LongRightArrow is called the projection of v &LongRightArrow perpendicular to w &LongRightArrow or normal to w &LongRightArrow . (By definition this projection has zero component in the direction of w &LongRightArrow , and is therefore normal to w &LongRightArrow .)

3.3 Express the square of the component of v &LongRightArrow in the direction of w &LongRightArrow in terms of dot products. Solution

3.4 Express the component of v &LongRightArrow perpendicular to w &LongRightArrow in terms of dot products. Solution

3.5 Write out ( v &LongRightArrow − w &LongRightArrow ) · ( v &LongRightArrow − w &LongRightArrow ) using the linearity of the dot product in each of its arguments. What famous law does this establish? Solution

3.6 Express the projection of v &LongRightArrow on w &LongRightArrow in terms of dot products and the vector w &LongRightArrow . Solution


It is pretty much simply a short way to notate both vector field operations by looking at $ abla$ as a vector operator by writing egin abla=left(frac,frac,frac ight) end in $mathbb^3$, or equivalently egin abla=frachat+frachat+frachat. end Performing this vector operator on a scalar field gives you the expression for that field's gradient, whereas applying it to a vector field via a dot product gives you the vector field's divergence (analogoulsy for the cross product, which gives you the field's curl instead).

It is important to note, however, that unlike with regular three-vectors, this expression for divergence is not commutative, as the $ abla$ operator is not a vector in $mathbb^3$.


11.3E: Exercises for The Dot Product - Mathematics

The next topic for discussion is that of the dot product. Let’s jump right into the definition of the dot product. Given the two vectors (vec a = leftlangle <,,> ight angle ) and (vec b = leftlangle <,,> ight angle ) the dot product is,

Sometimes the dot product is called the scalar product. The dot product is also an example of an inner product and so on occasion you may hear it called an inner product.

  1. (vec v = 5vec i - 8vec j,,,vec w = vec i + 2vec j)
  2. (vec a = leftlangle <0,3, - 7> ight angle ,,,vec b = leftlangle <2,3,1> ight angle )

Not much to do with these other than use the formula.

a (vec vcenterdot vec w = 5 - 16 = - 11)

b (vec acenterdot vec b = 0 + 9 - 7 = 2)

Here are some properties of the dot product.

Properties

The proofs of these properties are mostly “computational” proofs and so we’re only going to do a couple of them and leave the rest to you to prove.

Proof of (vec ucenterdot left( ight) = vec ucenterdot vec v + vec ucenterdot vec w)

We’ll start with the three vectors, (vec u = leftlangle <,, ldots ,> ight angle ), (vec v = leftlangle <,, ldots ,> ight angle ) and (vec w = leftlangle <,, ldots ,> ight angle ) and yes we did mean for these to each have (n) components. The theorem works for general vectors so we may as well do the proof for general vectors.

Now, as noted above this is pretty much just a “computational” proof. What that means is that we’ll compute the left side and then do some basic arithmetic on the result to show that we can make the left side look like the right side. Here is the work.

Proof of : If (vec vcenterdot vec v = 0) then (vec v = vec 0)

This is a pretty simple proof. Let’s start with (vec v = leftlangle <,, ldots ,> ight angle ) and compute the dot product.

Now, since we know (v_i^2 ge 0) for all (i) then the only way for this sum to be zero is to in fact have (v_i^2 = 0). This in turn however means that we must have ( = 0) and so we must have had (vec v = vec 0).

There is also a nice geometric interpretation to the dot product. First suppose that ( heta) is the angle between (vec a) and (vec b) such that (0 le heta le pi ) as shown in the image below.

We can then have the following theorem.

Theorem

Proof

Let’s give a modified version of the sketch above.

The three vectors above form the triangle AOB and note that the length of each side is nothing more than the magnitude of the vector forming that side.

The Law of Cosines tells us that,

Also using the properties of dot products we can write the left side as,

Our original equation is then,

The formula from this theorem is often used not to compute a dot product but instead to find the angle between two vectors. Note as well that while the sketch of the two vectors in the proof is for two dimensional vectors the theorem is valid for vectors of any dimension (as long as they have the same dimension of course).

Let’s see an example of this.

We will need the dot product as well as the magnitudes of each vector.

[vec acenterdot vec b = - 22hspace<0.25in>hspace<0.25in>left| ight| = sqrt <26>hspace<0.25in>hspace<0.25in>left| ight| = sqrt <29>]

The dot product gives us a very nice method for determining if two vectors are perpendicular and it will give another method for determining when two vectors are parallel. Note as well that often we will use the term orthogonal in place of perpendicular.

Now, if two vectors are orthogonal then we know that the angle between them is 90 degrees. From (eqref) this tells us that if two vectors are orthogonal then,

Likewise, if two vectors are parallel then the angle between them is either 0 degrees (pointing in the same direction) or 180 degrees (pointing in the opposite direction). Once again using (eqref) this would mean that one of the following would have to be true.

  1. (vec a = leftlangle <6, - 2, - 1> ight angle ,,,vec b = leftlangle <2,5,2> ight angle )
  2. (displaystyle vec u = 2vec i - vec j,,,vec v = - frac<1><2>vec i + frac<1><4>vec j)

First get the dot product to see if they are orthogonal.

[vec acenterdot vec b = 12 - 10 - 2 = 0]

The two vectors are orthogonal.

Again, let’s get the dot product first.

[vec ucenterdot vec v = - 1 - frac<1> <4>= - frac<5><4>]

So, they aren’t orthogonal. Let’s get the magnitudes and see if they are parallel.

[vec ucenterdot vec v = - frac<5> <4>= - sqrt 5 left( ><4>> ight) = - left| ight|,,left| ight|]

So, the two vectors are parallel.

There are several nice applications of the dot product as well that we should look at.

Projections

The best way to understand projections is to see a couple of sketches. So, given two vectors (vec a) and (vec b) we want to determine the projection of (vec b) onto (vec a). The projection is denoted by (< olimits> _>vec b). Here are a couple of sketches illustrating the projection.

So, to get the projection of (vec b) onto (vec a) we drop straight down from the end of (vec b)until we hit (and form a right angle) with the line that is parallel to (vec a). The projection is then the vector that is parallel to (vec a), starts at the same point both of the original vectors started at and ends where the dashed line hits the line parallel to (vec a).

There is a nice formula for finding the projection of (vec b) onto (vec a). Here it is,

Note that we also need to be very careful with notation here. The projection of (vec a) onto (vec b)is given by

We can see that this will be a totally different vector. This vector is parallel to (vec b), while (< olimits> _>vec b) is parallel to (vec a). So, be careful with notation and make sure you are finding the correct projection.

We need the dot product and the magnitude of (vec a).

For comparison purposes let’s do it the other way around as well.

We need the dot product and the magnitude of (vec b).

As we can see from the previous two examples the two projections are different so be careful.

Direction Cosines

This application of the dot product requires that we be in three dimensional space unlike all the other applications we’ve looked at to this point.

Let’s start with a vector, (vec a), in three dimensional space. This vector will form angles with the (x)-axis (a ), the (y)-axis (b ), and the (z)-axis (g ). These angles are called direction angles and the cosines of these angles are called direction cosines.

Here is a sketch of a vector and the direction angles.

The formulas for the direction cosines are,

where (vec i), (vec j) and (vec k) are the standard basis vectors.

Let’s verify the first dot product above. We’ll leave the rest to you to verify.

[vec acenterdot ,vec i = leftlangle <,,> ight angle centerdot leftlangle <1,0,0> ight angle = ]

Here are a couple of nice facts about the direction cosines.

  1. The vector (vec u = leftlangle ight angle ) is a unit vector.


11.3E: Exercises for The Dot Product - Mathematics

The dot product of two vectors A and B is a key operation in using vectors in geometry.

In the coordinate space of any dimension (we will be mostly interested in dimension 2 or 3):

Examples: Let A = (1, 2, -1), B = (3, 2, 1), C = (0, -5, 2). Note that if we set D = 2B - C, then D = (6, 9, 0).

A . B = 1*3 + 2*2 + (-1)*1 = 6

A . C = 1*0 + 2*(-5) + (-1)*2 = -12

A . D = 1*6 + 2*9 + (-1)*0 = 24

Notice that A . D = A . (2B - C) = 24 and also 2A . B - A . C = 2*6 - (-12) = 24. This is not a coincidence, but follows from the algebraic properties of the dot product..

Algebraic Properties of the Dot Product

These properties are extremely important, though they are a little boring to prove. It takes a second look to see that anything is going on at all, but look twice or 3 times.

(1) (Commutative Property) For any two vectors A and B, A . B = B . A.

(2) (Scalar Multiplication Property) For any two vectors A and B and any real number c, (cA) . B = A . (cB) = c(A . B)

(3) (Distributive Property) For any 3 vectors A, B and C, A . (B+C) = A . B + A . C.

Let A, B, C, D be as above for the next 3 exercises.

Exercise 1: Compute B . A and compare with A . B. Can you see why these numbers are the same in this example and will always be the same for any choice of A and B?

Exercise 2: Let c = 10. Write down 10A and 10B. Then compute each of the 3 terms in property (2) above and see that they are in fact the same. Again look to see why this is true.

Exercise 3: Compute E = B+C. Check that A . E is really the sum of A . B and A . C.

Exercise 4. Show how you can combine (3) with (2) in several steps to show that for any vectors A, B, C and any real numbers h and k, A . (hB+kC) = A . hB + A . kC. Tell how this explains why for the special A, B, C, D above, A . D = A . (2B - C).

Exercise 5: Use the properties above to expand (aA + bC) . (cC + dD) into a sum of four terms, starting with ac(A . C) + .

Exercise 6: For a point A = (3, 4) in the plane, compute the square root of A . A. Tell why this number is the distance from A to O.

Geometric Properties of the Dot Product

Length and Distance Formula

For A = (a1, a2, . an), the dot product A . A is simply the sum of squares of each entry.

In the plane or 3-space, the Pythagorean theorem tells us that the distance from O to A, which we think of as the length of vector OA, (or just length of A), is the square root of this number.

Definition. For A in n-space, the length of A = square root of A . A. This length is written |A|. so |A| 2 = A . A.

Likewise, the Pythagorean theorem also shows that the distance from A to B is the length of AB, which is the length of B-A.

Definition: For A and B in n-space, the distance from A to B is the length |B-A|.
Note: This equals |A-B| also.

Law of Cosines

Cosine Theorem: For A and B in the plane or in space, A . B = |A| |B| cos AOB

In the triangle AOB, if we let the length of side AB = c, then by definition of distance,

Proof: From the algebraic properties,

|A - B| 2 = A . B - B . A - B . B = |A| 2 + |B| 2 - 2A . B.

But from geometry, since |A| = |OA| = side opposite B, etc., the Law of Cosines for triangle AOB says

c 2 = |A| 2 + |B| 2 - 2|A| |B| cos AOB.

Comparing these, all but one term is the same in each, so we see that A . B = |A| |B| cos AOB

Exercise: Compute the angle between (1, 1, 1) and (0, 0, 1).

Exercise: In the plane, let A = (1, 2), B = (3, 4), C = (-2, -1). Use the dot product to compute all the side lengths and all the angles of this triangle.

Exercise: In the plane, let A = (1, 2, 1), B = (3, 4, 1), C = (-2, -1, 3). Use the dot product to compute all the side lengths and all the angles of this triangle.

Orthogonal Vectors

The cosine of a right angle = 0, so a very important special case of the cosine theorem is this:

Orthogonal Vector Theorem: Two vectors A and B are orhthogonal if and only if their dot product is zero.


Test: Dot Product And Cross Product

Determine the magnitude of the projection of the vector force F = 100N, onto the u axis, from the figure given below.

The component of the force in the u axis, it is equal to 100cos(45˚). We need to consider the triangle and then accordingly apply the trigonometry. This is one of the way of resolving the components.

For two vectors A and B, what is A.B (if they have angle &alpha between them)?

The dot product of the two vectors is always the product of the magnitudes of the two forces and the cosine of the angle between them. We need to consider the triangle and then accordingly apply the trigonometry. This is one of the way of resolving the components.

Multiplicative law: a(A.B) = Ax(aB)

Multiplicative law: A.(B+D) = (A.B) + (A.D)

For three vectors A, B and D the various laws are. Communitive law: A.B =B.A. While distributive law is A.(B+D) = (A.B) + (A.D). And multiplication law is a(A.B) = A.(aB).

For three vectors A, B and D the various laws are. Communitive law: A.B =B.A. While distributive law is A.(B+D) = (A.B) + (A.D). And multiplication law is a(A.B) = A.(aB).

What is multiplication law?

For three vectors A, B and D the various laws are. Communitive law: A.B =B.A. While distributive law is A.(B+D) = (A.B) + (A.D). And multiplication law is a(A.B) = A.(aB).

Which is true for two vector A = A1i + A2j + A3k and B = B1i + B2j + B3k?

The multiplication of x, y and z components with their respective same component give a scalar, equal to 1, i.e. i.i = 1 and j.j = 1, while jxj =0. This is the basic principle of the vector algebra which needs to apply wherever needed.

Determine the magnitude of the force F = 300j parallel to the direction of AB?

Force component in the direction parallel to the AB is given by unit vector 0.286i + 0.857j + 0.429k. Now (300j).(0.286i + 0.857j + 0.429k) = 257.1N. Just try to resolve the force into it&rsquos particular components.

As the dot product of only the same Cartesian component is unity, i.e. i.i = 1 and j.j =1, rest all remaining dot product will give 0(i.j = 0 and j.k = 0). Cross product of the same plane vectors always give zero. And dot product of the same plane vector gives a scalar quantity.

<(i.i) + (-i.j) + (-k.k) + (k.i)>= 0. As (1 + 0 + 0 &ndash 1). Cross product of the same plane vectors always give zero. And dot product of the same plane vector gives a scalar quantity.

&alpha = cos-1(A.B/AB). What is the range of &alpha?

Cosine inverse function is defined only between 0˚ to 180˚. It cannot be defined under any of the given range, because this is the principle range of the inverse cosine function.

What is the value of <(AxB).(C.D)x(AxB).(C.D)>? For the four vectors A, B, C and D, with A, B, C and D all lying in the same plane?

Here we are trying to do the cross product of the two vectors in the same plane. Which will give us zero. While dot product of the same plane vector will give a scalar quantity, not zero.

What is (AxB).(BxA) or A = A1i + A2j + A3k and B = B1i + B2j + B3k?

A1B1A2B2i + A2B2A3B3j + A3B3A1B1k

A1B1A1B2i + A2B2A3B2j + A3B3A1B3k

A1B1A2B1i + A2B2A2B3j + A3B3A2B1k

Here we are trying to do the cross product of the two vectors in the same plane. Which will give us zero. While dot product of the same plane vector will give a scalar quantity, not zero.

What is the value of &theta in the figure given below ?

Fab= -(3/5)j + (4/5)k F= (4/5)i &ndash (3/5)j &theta = cos-1(Fab.F) = 69˚. This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy.

Refer to the diagram given below, determine the magnitude of the force F = 300j perpendicular to the direction of AB.

Force&rsquos component in the direction perpendicular to AB is -73.5i + 80j &ndash 110k N. This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy.

Determine the magnitude of the projection of the vector force F = 100N, onto the v axis, from the figure given below.

The component of the force in the v axis, it is equal to 100cos(15˚). This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy.

What is the dot product of two vectors which are having magnitude equal to unity and are making an angle of 45°?

The dot product of two vectors having the angle between them equal to 45° will have the product of the vector&rsquos magnitude. As the vectors are of unit magnitude, their product will be unity. Thus the magnitude factor would be cosine function at 45°.

Mathematically, for two vectors A and B of any magnitude, the cross product of both, i.e. AxB = given by:

The cross product of two vectors gives a vector which is perpendicular to both of the vectors. And the mathematic equation for the same is given by: |A||B|sinØ. And the dot product of the same by any of the other two vectors will give the answer zero, as perpendicular.

Commutative law is valid for the cross product of two vectors. (Commutative law: PxQ = QxP for two vectors P and Q)

This statement is wrong. It is not possible, unless we apply a negative sign to the RHS of the equation. That is PxQ = -(QxP). It is because, if you curl your wrist from one vector towards another vector, the thumb projected will give the direction of the cross product. Thus if you reverse the direction, negative sign is necessary.

Which among the following is the distributive law for the cross product of three vectors?

The distributive law works just like the simple multiplication of the constant before the brackets. That is in the equation Px(Q+S) = (PxQ) + (PxS), P is crossed by Q and S both. This is simple as, if we first add the two vectors and then do the cross product or we first do the cross product and the do the sum.

Which statement is true? (For three vectors P, Q and R)

Associative law for cross product: (PxQ)xS = Px(QxS)

Associative law for cross product: (PxQ)xS &ne Px(QxS)

Associative law for cross product: (PxQ)xS > Px(QxS)

Associative law for cross product: (PxQ)xS < Px(QxS)

The associative law is defined in the cross product of three vectors. This property is though valid for the dot product. But for the cross product, it is not true. It is because, in the dot product the final result is the scalar quantity, but in cross product it is the direction too. Thus the answer.

Which of the following is true?

As the mathematic equation for the cross product is having a cosine function in it, in which the angle used in the function is the angle between the vectors. Thus the cross product will be zero if the angle between them is 90.

Which of them is not correct?

As asked, the one which is not correct is the third one. The product is containing the cosine function, and the angle which is going to be inserted in the function is the angle between the vectors. Thus if the angle is 90, then the cross will be zero.

The ___________ forces do not cause the rotation.

The concurrent forces are the which are somewhere touching the axis of rotation. If any of the force is touching that axis, that force is not considered, or is insufficient to cause a rotation. If a force is concurrent then the perpendicular distance of the force from the line of axis is zero, thus no rotation. As we know rotation is caused by moment.

The tendency of a force to rotate the body is called the moment of the force.

The moment of the force about a axis or a point gives the measure of the tendency of the force. It is the cause of the body&rsquos rotation, about that point or at that axis. Thus, the tendency of a force to rotate the body is called the moment of the force.


Running the Code

Since we stepped through the build.sh and run.sh files in the Hello World example, we will not do so here. However, we will call to your attention some key differences in each file.

In the build file, first notice that we are referring to internal.ldf instead of fast.ldf . Linker Descriptor Files (LDF)s help you choose the memory layout of your Epiphany application.

  • If you are planning on storing everything external to the Epiphany chip, use legacy.ldf .
  • If you plan to have some memory outside (like in the case of our Hello World application, where we used some SDRAM memory, use fast.ldf .
  • If all the memory (like in this application) is internal to the epiphany chip, use internal.ldf .