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11.2E: Exercises for Vectors in Space


1) Consider a rectangular box with one of the vertices at the origin, as shown in the following figure. If point (A(2,3,5)) is the opposite vertex to the origin, then find

a. the coordinates of the other six vertices of the box and

b. the length of the diagonal of the box determined by the vertices (O) and (A).

Answer:
a. ((2,0,5),(2,0,0),(2,3,0),(0,3,0),(0,3,5),(0,0,5)) b. (sqrt{38})

2) Find the coordinates of point (P) and determine its distance to the origin.

For exercises 3-6, describe and graph the set of points that satisfies the given equation.

3) ((y−5)(z−6)=0)

Answer:
A union of two planes: (y=5) (a plane parallel to the (xz)-plane) and (z=6) (a plane parallel to the (xy)-plane)

4) ((z−2)(z−5)=0)

5) ((y−1)^2+(z−1)^2=1)

Answer:
A cylinder of radius (1) centered on the line (y=1,z=1)

6) ((x−2)^2+(z−5)^2=4)

7) Write the equation of the plane passing through point ((1,1,1)) that is parallel to the (xy)-plane.

Answer:
(z=1)

8) Write the equation of the plane passing through point ((1,−3,2)) that is parallel to the (xz)-plane.

9) Find an equation of the plane passing through points ((1,−3,−2), (0,3,−2),) and ((1,0,−2).)

Answer:
(z=−2)

10) Find an equation of the plane passing through points ((1,9,2), (1,3,6),) and ((1,−7,8).)

For exercises 11-14, find the equation of the sphere in standard form that satisfies the given conditions.

11) Center (C(−1,7,4)) and radius (4)

Answer:
((x+1)^2+(y−7)^2+(z−4)^2=16)

12) Center (C(−4,7,2)) and radius (6)

13) Diameter (PQ,) where (P(−1,5,7)) and (Q(−5,2,9))

Answer:
(x+3)^2+(y−3.5)^2+(z−8)^2=dfrac{29}{4})

14) Diameter (PQ,) where (P(−16,−3,9)) and (Q(−2,3,5))

For exercises 15 and 16, find the center and radius of the sphere with an equation in general form that is given.

15) ( x^2+y^2+z^2−4z+3=0)

Answer:
Center (C(0,0,2)) and radius (1)

16) (x^2+y^2+z^2−6x+8y−10z+25=0)

For exercises 17-20, express vector ( vecd{PQ} ) with the initial point at (P) and the terminal point at (Q)

(a.) in component form and

(b.) by using standard unit vectors.

17) (P(3,0,2)) and (Q(−1,−1,4))

Answer:
(a. vecd{PQ}=⟨−4,−1,2⟩)
(b. vecd{PQ}=−4hat{mathbf i}−hat{mathbf j}+2hat{mathbf k})

18) (P(0,10,5)) and (Q(1,1,−3))

19) (P(−2,5,−8)) and (M(1,−7,4)), where (M) is the midpoint of the line segment (overline{PQ})

Answer:
(a. vecd{PQ}=⟨6,−24,24⟩)
(b. vecd{PQ}=6hat{mathbf i}−24hat{mathbf j}+24hat{mathbf k})

20) (Q(0,7,−6)) and (M(−1,3,2)), where (M) is the midpoint of the line segment (overline{PQ})

21) Find terminal point (Q) of vector (vecd{PQ}=⟨7,−1,3⟩) with the initial point at (P(−2,3,5).)

Answer:
(Q(5,2,8))

22) Find initial point (P) of vector (vecd{PQ}=⟨−9,1,2⟩) with the terminal point at (Q(10,0,−1).)

For exercises 23-26, use the given vectors (vecs a) and (vecs b) to find and express the vectors (vecs a+vecs b, ,4vecs a), and (−5vecs a+3vecs b) in component form.

23) (quad vecs a=⟨−1,−2,4⟩,quad vecs b=⟨−5,6,−7⟩)

Answer:
(vecs a+vecs b=⟨−6,4,−3⟩, 4vecs a=⟨−4,−8,16⟩, −5vecs a+3vecs b=⟨−10,28,−41⟩)

24) (quad vecs a=⟨3,−2,4⟩,quad vecs b=⟨−5,6,−9⟩)

25) (quad vecs a=−hat{mathbf k},quad vecs b=−hat{mathbf i})

Answer:
(vecs a+vecs b=⟨−1,0,−1⟩, 4vecs a=⟨0,0,−4⟩, −5vecs a+3vecs b=⟨−3,0,5⟩)

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

For exercises 27-30, vectors (vecs u) and (vecs v) are given. Find the magnitudes of vectors (vecs u−vecs v) and (−2vecs u).

27) (quad vecs u=2hat{mathbf i}+3hat{mathbf j}+4hat{mathbf k}, quad vecs v=−hat{mathbf i}+5hat{mathbf j}−hat{mathbf k})

Answer:
(|vecs u−vecs v|=sqrt{38}, quad |−2vecs u|=2sqrt{29})

28) (quad vecs u=hat{mathbf i}+hat{mathbf j}, quad vecs v=hat{mathbf j}−hat{mathbf k})

29) (quad vecs u=⟨2cos t,−2sin t,3⟩, quad vecs v=⟨0,0,3⟩,quad) where (t) is a real number.

Answer:
(|vecs u−vecs v|=2, quad |−2vecs u|=2sqrt{13})

30) (quad vecs u=⟨0,1,sinh t⟩, quad vecs v=⟨1,1,0⟩,quad) where (t) is a real number.

For exercises 31-36, find the unit vector in the direction of the given vector ( vecs a) and express it using standard unit vectors.

31) (quad vecs a=3hat{mathbf i}−4hat{mathbf j})

Answer:
(frac{3}{5}hat{mathbf i}−frac{4}{5}hat{mathbf j})

32) (quad vecs a=⟨4,−3,6⟩)

33) (quad vecs a=vecd{PQ}), where ( P(−2,3,1)) and (Q(0,−4,4))

Answer:
(frac{sqrt{62}}{31}hat{mathbf i}−frac{7sqrt{62}}{62}hat{mathbf j}+frac{3sqrt{62}}{62}hat{mathbf k})

34) (quad vecs a=vecd{OP},) where (P(−1,−1,1))

35) (quad vecs a=vecs u−vecs v+vecs w,) where (vecs u=hat{mathbf i}−hat{mathbf j}−hat{mathbf k},quad vecs v=2hat{mathbf i}−hat{mathbf j}+hat{mathbf k}, quad) and (vecs w=−hat{mathbf i}+hat{mathbf j}+3hat{mathbf k})

Answer:
(−frac{sqrt{6}}{3}hat{mathbf i}+frac{sqrt{6}}{6}hat{mathbf j}+frac{sqrt{6}}{6}hat{mathbf k})

36) (quad vecs a=2vecs u+vecs v−vecs w,quad) where ( vecs u=hat{mathbf i}−hat{mathbf k}, quad vecs v=2hat{mathbf j} quad), and ( vecs w=hat{mathbf i}−hat{mathbf j})

37) Determine whether (vecd{AB}) and (vecd{PQ}) are equivalent vectors, where (A(1,1,1),,B(3,3,3),,P(1,4,5),) and (Q(3,6,7).)

Answer:
Equivalent vectors

38) Determine whether the vectors (vecd{AB}) and (vecd{PQ}) are equivalent, where ( A(1,4,1),, B(−2,2,0),, P(2,5,7),) and ( Q(−3,2,1)).

For exercises 39-42, find vector ( vecs u) with a magnitude that is given and satisfies the given conditions.

39) (quad vecs v=⟨7,−1,3⟩, , ‖vecs u‖=10), and (vecs u) and (vecs v) have the same direction

Answer:
(vecs u=⟨frac{70sqrt{59}}{59},−frac{10sqrt{59}}{59},frac{30sqrt{59}}{59}⟩)

40) (quad vecs v=⟨2,4,1⟩,, ‖vecs u‖=15), and (vecs u) and (vecs v) have the same direction

41) (quad vecs v=⟨2sin t,, 2cos t,1⟩, ‖vecs u‖=2,vecs u) and (vecs v) have opposite directions for any (t), where (t) is a real number

Answer:
(vecs u=⟨−frac{4sqrt{5}}{5}sin t,−frac{4sqrt{5}}{5}cos t,−frac{2sqrt{5}}{5}⟩)

42) (quad vecs v=⟨3sinh t,0,3⟩,, ‖vecs u‖=5), and (vecs u) and (vecs v) have opposite directions for any (t), where (t) is a real number

43) Determine a vector of magnitude (5) in the direction of vector (vecd{AB}), where (A(2,1,5)) and (B(3,4,−7).)

Answer:
(⟨frac{5sqrt{154}}{154},frac{15sqrt{154}}{154},−frac{30sqrt{154}}{77}⟩)

44) Find a vector of magnitude (2) that points in the opposite direction than vector (vecd{AB}), where (A(−1,−1,1)) and (B(0,1,1).) Express the answer in component form.

45) Consider the points (A(2,α,0), , B(0,1,β),) and (C(1,1,β)), where (α) and (β) are negative real numbers. Find (α) and (β) such that (|vecd{OA}−vecd{OB}+vecd{OC}|=|vecd{OB}|=4.)

Answer:
(α=−sqrt{7}, ,β=−sqrt{15})

46) Consider points (A(α,0,0),,B(0,β,0),) and (C(α,β,β),) where (α) and (β) are positive real numbers. Find (α) and (β) such that (|overline{OA}+overline{OB}|=sqrt{2}) and (|overline{OC}|=sqrt{3}).

47) Let (P(x,y,z)) be a point situated at an equal distance from points (A(1,−1,0)) and (B(−1,2,1)). Show that point (P) lies on the plane of equation (−2x+3y+z=2.)

48) Let (P(x,y,z)) be a point situated at an equal distance from the origin and point (A(4,1,2)). Show that the coordinates of point P satisfy the equation (8x+2y+4z=21.)

49) The points (A,B,) and (C) are collinear (in this order) if the relation ({|vecd{AB}|+|vecd{BC}|=|vecd{AC}|}) is satisfied. Show that (A(5,3,−1),, B(−5,−3,1),) and (C(−15,−9,3)) are collinear points.

50) Show that points (A(1,0,1), , B(0,1,1),) and (C(1,1,1)) are not collinear.

51) [T] A force (vecs F) of (50 ,N) acts on a particle in the direction of the vector (vecd{OP}), where (P(3,4,0).)

a. Express the force as a vector in component form.

b. Find the angle between force (vecs F) and the positive direction of the (x)-axis. Express the answer in degrees rounded to the nearest integer.

Answer:
(a. vecs F=⟨30,40,0⟩; quad b. 53°)

52) [T] A force (vecs F) of (40,N) acts on a box in the direction of the vector (vecd{OP}), where (P(1,0,2).)

a. Express the force as a vector by using standard unit vectors.

b. Find the angle between force (vecs F) and the positive direction of the (x)-axis.

53) If (vecs F) is a force that moves an object from point (P_1(x_1,y_1,z_1)) to another point (P_2(x_2,y_2,z_2)), then the displacement vector is defined as (D=(x_2−x_1)hat{mathbf i}+(y_2−y_1)hat{mathbf j}+(z_2−z_1)hat{mathbf k}). A metal container is lifted (10) m vertically by a constant force (vecs F). Express the displacement vector (D) by using standard unit vectors.

Answer:
(vecs D=10hat{mathbf k})

54) A box is pulled (4) yd horizontally in the (x)-direction by a constant force ( vecs F). Find the displacement vector in component form.

55) The sum of the forces acting on an object is called the resultant or net force. An object is said to be in static equilibrium if the resultant force of the forces that act on it is zero. Let (vecs F_1=⟨10,6,3⟩, vecs F_2=⟨0,4,9⟩), and (vecs F_3=⟨10,−3,−9⟩) be three forces acting on a box. Find the force (vecs F_4) acting on the box such that the box is in static equilibrium. Express the answer in component form.

Answer:
(vecs F_4=⟨−20,−7,−3⟩)

56) [T] Let (vecs F_k=⟨1,k,k^2⟩, k=1,...,n) be (n) forces acting on a particle, with (n≥2.)

a. Find the net force (vecs F=sum_{k=1}^nvecs F_k.) Express the answer using standard unit vectors.

b. Use a computer algebra system (CAS) to find (n) such that (|vecs F|<100.)

57) The force of gravity ( vecs F) acting on an object is given by ( vecs F=mvecs g), where (m) is the mass of the object (expressed in kilograms) and (vecs g) is acceleration resulting from gravity, with ( |vecs g|=9.8 ,N/kg.) A 2-kg disco ball hangs by a chain from the ceiling of a room.

a. Find the force of gravity (vecs F) acting on the disco ball and find its magnitude.

b. Find the force of tension (vecs T) in the chain and its magnitude.

Express the answers using standard unit vectors.

Answer:
(a. vecs F=−19.6hat{mathbf k}, quad |vecs F|=19.6 ,N)
(b. vecs T=19.6hat{mathbf k}, quad |vecs T|=19.6 ,N)

58) A 5-kg pendant chandelier is designed such that the alabaster bowl is held by four chains of equal length, as shown in the following figure.

a. Find the magnitude of the force of gravity acting on the chandelier.

b. Find the magnitudes of the forces of tension for each of the four chains (assume chains are essentially vertical).

59) [T] A 30-kg block of cement is suspended by three cables of equal length that are anchored at points (P(−2,0,0), Q(1,sqrt{3},0),) and (R(1,−sqrt{3},0)). The load is located at (S(0,0,−2sqrt{3})), as shown in the following figure. Let (vecs F_1, vecs F_2), and (vecs F_3) be the forces of tension resulting from the load in cables (RS,QS,) and (PS,) respectively.

a. Find the gravitational force (vecs F) acting on the block of cement that counterbalances the sum (vecs F_1+vecs F_2+vecs F_3) of the forces of tension in the cables.

b. Find forces (vecs F_1, vecs F_2,) and ( vecs F_3). Express the answer in component form.

Answer:
a. (vecs F=−294hat{mathbf k}) N;
b. (vecs F_1=⟨−frac{49sqrt{3}}{3},49,−98⟩, vecs F_2=⟨−frac{49sqrt{3}}{3},−49,−98⟩), and (vecs F_3=⟨frac{98sqrt{3}}{3},0,−98⟩) (each component is expressed in newtons)

60) Two soccer players are practicing for an upcoming game. One of them runs 10 m from point A to point B. She then turns left at (90°) and runs 10 m until she reaches point C. Then she kicks the ball with a speed of 10 m/sec at an upward angle of (45°) to her teammate, who is located at point A. Write the velocity of the ball in component form.

61) Let (vecs r(t)=⟨x(t),, y(t), , z(t)⟩) be the position vector of a particle at the time (t∈[0,T]), where (x,y,) and (z) are smooth functions on ([0,T]). The instantaneous velocity of the particle at time (t) is defined by vector (vecs v(t)=⟨x'(t), , y'(t), , z'(t)⟩), with components that are the derivatives with respect to (t), of the functions (x, y), and (z), respectively. The magnitude (∥vecs v(t)∥) of the instantaneous velocity vector is called the speed of the particle at time (t). Vector (vecs a(t)=⟨x''(t), , y''(t), , z''(t)⟩), with components that are the second derivatives with respect to (t), of the functions (x,y,) and (z), respectively, gives the acceleration of the particle at time (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 is expressed in seconds.

a. Find the instantaneous velocity, speed, and acceleration of the particle after the first second. Round your answer to two decimal places.

b. Use a CAS to visualize the path of the particle—that is, the set of all points of coordinates ((cos t,sin t,2t),) where (t∈[0,30].)

Answer:
(a. vecs v(1)=⟨−0.84,0.54,2⟩) (each component is expressed in centimeters per second); (∥vecs v(1)∥=2.24) (expressed in centimeters per second); (vecs a(1)=⟨−0.54,−0.84,0⟩) (each component expressed in centimeters per second squared);

(b.)

62) [T] Let (vecs r(t)=⟨t,2t^2,4t^2⟩) be the position vector of a particle at time (t) (in seconds), where (t∈[0,10]) (here the components of (vecs r) are expressed in centimeters).

a. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Use a CAS to visualize the path of the particle defined by the points ((t, , 2t^2, , 4t^2),) where (t∈[0, , 60].)

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.


Vectors in space: how to graph, applications, exercises

A vector in space is everyone represented by a coordinate system given by x, Y Y z. Almost always the plane xy is the plane of the horizontal surface and the axis z represents height (or depth).

The Cartesian coordinate axes shown in figure 1 divide space into 8 regions called octants, analogous to how axes xY divide the plane into 4 quadrants. We will then have 1st octant, 2nd octant and so on.

Figure 1 contains a representation of a vector v in the space. Some perspective is required to create the illusion of three dimensions on the plane of the screen, which is achieved by drawing an oblique view.

To graph a 3D vector, one must use the dotted lines that determine the coordinates of the projection or "shadow" on the grid. v Over the surface x-y. This projection begins at O ​​and ends at the green point.

Once there, you must continue along the vertical to the necessary height (or depth) according to the value of z, until reaching P. The vector is drawn starting from O and ending at P, which in the example is in the 1st octant.


Grid supplied three phase voltage

A grid supplied three phase voltage will look like the animation below. This is what an induction machine is experiencing when connected directly to the grid.

The three phased system is illustrated using two different, but equal forms:

A vector diagram showing all three phases and their vector sum (space vector).

An ordinary instantaneous sine wave representation, also showing the resultant space vector.

An ordinary three phased system, here shown in both vector form and in sinusoidal form. The black vector is the resultant space vector a vector sum obtained by adding the three vectors. As can be seen, the space vector's magnitude is always constant.


Constructing vectors

Vectors in two dimensions have a type like V2 n , where n is some numeric type of scalar values (often Double ). (One can also work with other vector spaces with any number of dimensions in this tutorial we'll stick to the 2D case.)

The first thing to learn is how to create values of type V2 n . There are many options:

zero is the zero vector, that is, the vector with zero magnitude (and no direction, or perhaps every direction). zero is rarely useful on its own, but can come in handy e.g. as an argument to a function expecting a vector input.

unitX and unitY are the length-one vectors in the positive (x) and (y) directions, respectively. To create a length-(l) vector you can apply scaling to unitX or unitY , like unitX # scale 3 or 3 *^ unitX (see Vector operations).

Also, unit_X and unit_Y are like unitX and unitY but point in the corresponding negative directions.

To create a vector with given (x)- and (y)- components, you can use the function r2 :: (n, n) -> V2 n :

As you can see, r2 is especially useful if you already have pairs representing vector components (which is not uncommon if the components are coming from some other data source).

You can also use the data constructor V2 :

You can also use ( ^& ) to construct vector literals, like so:

This can make for convenient and pleasant notation. However, it does have some drawbacks, namely:

( ^& ) is extremely general so its type is unhelpful.

Related to the above, literal vector expressions like 1 ^& 2 must be used in a context where the type can be inferred (or else a type annotation must be added). This is because (as we will see later) ( ^& ) can also be used to construct points as well as higher-dimensional vectors.

Only you can decide whether the tradeoffs are worth it in a given situation.

You can construct vectors from Direction s using the fromDirection function. fromDirection takes a Direction and constructs a unit (i.e. magnitude 1) vector pointing in the given direction.

One final way to construct vectors is using the function e . By definition, e a == unitX # rotate a , but sometimes calling e can be more convenient. The name e is a sort of pun: in the same way that a complex number with magnitude (r) and angle ( heta) can be constructed as (r e^), a vector with given magnitude and direction can be constructed as r *^ e theta . (Note that e is not exported from Diagrams.Prelude if you wish to use it you must import it from Diagrams.TwoD.Vector .)

Construct each of the following images.

The circles have radius 1, and are arranged in the shape of a radius-5 semicircle.

30 spokes with lengths 1, 2, and 3.


Exercise 3.12 Derivatives of Killing vectors

Question

Answers

Wilhelm Karl Joseph Killing 1847-1923
The text gives a few clues about how to get from (1) to (2). What we know about Killing vectors is that they satisfy Killing's equation
egin
abla_<(mu>K_< u)>equivfrac<1><2>left( abla_mu K_ u+ abla_ u K_mu ight)=0&phantom <10000>(3) onumber
Rightarrow abla_mu K_ u=- abla_ u K_mu&phantom <10000>(4) onumber
end## abla_mu K_ u## is antisymmetric.

The Killing vectors in (1) and (2) are given in contravariant form.
The Riemann tensor is antisymmetric* in its last two indices like (7) and has other interesting symmetries, not all independent, e.g. (6) comes form (7) and (8):
egin
&R_< hosigmamu u>=g_< au ho>R_< sigma umu>^ au&phantom <10000>(5) onumber
ext

As I found out after much effort, it is also important to remember the definition of the Riemann tensor when the connection is torsion free. It measures the difference between taking covariant derivatives of a vector going the two opposite ways round a path (Carroll 3.112)
egin
left[ abla_ ho, abla_sigma ight]X^mu=R_< usigma ho>^mu X^ u&phantom <10000>(16) onumber
endThat equation is pretty similar to (1) and then there's (4).

The second part is very simple if you remember that if a Killing vector exists then it is always possible to find a coordinate system where it is one of the basis vectors and the metric will be independent of the coordinate for that basis vector. Carroll stated this after the Killing equation at his 3.174.

I was unable to answer the questions but was guided by a solution I stumbled across on Semantic Scholar by Professor Alan Guth.


11.2E: Exercises for Vectors in Space

Finding Velocity and Acceleration Vectors in Space In Exercises 11-20, the position vector r describes the path or an object moving in space.

(a) Kind the velocity vector, speed, and acceleration vector of the object.

(b) Evaluate the velocity vector and acceleration vector of the object at the given value of t.

Position Vector Time

Finding Velocity and Acceleration Vectors in Space In Exercises 11-20, the position vector r describes the path or an object moving in space.

(a) Kind the velocity vector, speed, and acceleration vector of the object.

(b) Evaluate the velocity vector and acceleration vector of the object at the given value of t.

Position Vector Time

Finding Velocity and Acceleration Vectors in Space In Exercises 11-20, the position vector r describes the path or an object moving in space.

(a) Kind the velocity vector, speed, and acceleration vector of the object.

(b) Evaluate the velocity vector and acceleration vector of the object at the given value of t.


Cross Product

The cross product is only meaningful for 3D vectors. It takes two 3D vectors as input and returns another 3D vector as its result.

The result vector is perpendicular to the two input vectors. You can use the “left hand rule” to remember the direction of the output vector from the ordering of the input vectors. If the first parameter is matched up to the thumb of the hand and the second parameter to the forefinger, then the result will point in the direction of the middle finger. If the order of the parameters is reversed then the resulting vector will point in the exact opposite direction but will have the same magnitude.

The magnitude of the result is equal to the magnitudes of the input vectors multiplied together and then that value multiplied by the sine of the angle between them. Some useful values of the sine function are shown below:-

The cross product can seem complicated since it combines several useful pieces of information in its return value. However, like the dot product, it is very efficient mathematically and can be used to optimize code that would otherwise depend on slower transcendental functions such as sine and cosine.


5 Answers 5

Two sets can span the same subspace even if one is dependent and the other is not.

To solve such problems you can try, for example, seeing if every element of $S_2$ is a linear combination of elements from $S_1$ and vice verse. If they are, the sets span the same space.

Two sets of vectors in the same vector space, $S_1$ and $S_2$, span the same subspace if and only if:

  • Each vector in $S_1$ can be written as a linear combination of the vectors in $S_2$ and
  • Each vector in $S_2$ can be written as a linear combination of the vectors in $S_1$.

There may be ways of infering these properties without actually showing them directly (such as dimension arguments liike user6312 suggests), but it really ends up coming down to showing this holds or that this cannot hold.

Note that these conditions are not intrinsic: it's almost never enough to just look at $S_1$ without thinking about $S_2$, and to then look at $S_2$ without looking at $S_1$ it is only in very extreme circumstances that this suffices (when you can prove that the "sizes" of the spans don't match, a dimension argument) that this can prove to be enough.

So just figuring out if the sets are linearly dependent or independent is not enough in this case.

So: are $(-2,-6,0)$ and $(1,1,-2)$ each linear combinations of $(1,2,-1)$, $(0,1,1)$, and $(2,5,-1)$? Yes: you can try solving the two systems of linear equations: $egin alpha_1 left(egin12-1end ight) + eta_1left(egin011end ight) + gamma_1left(egin25-1end ight) &= left(egin-2-6end ight) alpha_2 left(egin12-1end ight) + eta_2left(egin011end ight) + gamma_2left(egin25-1end ight) &= left(egin11-2end ight) end$ and see if they each have solutions. (It can even be done both at once, by doing row reduction on $left(egin 1 & 0 & 2 & -2 & 1 2 & 1 & 5 & -6 & 1 -1 & 1 & -1 & 0 & -2 end ight).$ If either system has no solutions, then you know that not every vector in $S_2$ is in the span of $S_1$ and you are done if both systems have solutions, then every vector in $S_2$ is in the span of $S_1$, so $mathrm(S_2)subseteq mathrm(S_1)$.

Then you need to see if the converse inclusion holds: is every vector in $S_1$ a linear combination of the vectors in $S_2$? That is, can we solve the three systems of linear equations? $egin ho_1left(egin-2-6end ight) + sigma_1 left(egin11-2end ight) &= left(egin12-1end ight) ho_2left(egin-2-6end ight) + sigma_2 left(egin11-2end ight) &= left(egin011end ight) ho_3left(egin-2-6end ight) + sigma_3 left(egin11-2end ight) &= left(egin25-1end ight) end$ If we can solve all, then each vector in $S_1$ is in the span of $S_2$, so $mathrm(S_1)subseteq mathrm(S_2)$ together with the previous inclusion, this would show the spans are equal. If some equation cannot be solved, then not every vector in $S_1$ is in the span of $S_2$, so the spans are different.

(There is one thing that you can rescue from your efforts: since you proved that the set $S_1$ is linearly dependent, you can extract from it a linearly independent set (in this case, for instance, the first two vectors), and replace $S_1$ with that set of two vectors (because the third vector is a linear combination of the first two). That will mean that checking that "every vector in $S_1$ is a linear combination of the vectors in $S_2$" and checking that "every vector in $S_2$ is a linear combination of the vectors in $S_1$" will be simpler: instead of checking five things, you only need to check four.)

For this kind of problem, finding the reduced row echelon form (rref) seems to be a rather straightforward method.

For the set $S_1$, you form the matrix $ egin 1 & 2 &-1 0 & 1 & 1 2 & 5 &-1 endsim egin 1 & 2 &-1 0 & 1 & 1 0 & 1 & 1 endsim egin 1 & 2 &-1 0 & 1 & 1 0 & 0 & 0 endsim egin 1 & 0 &-3 0 & 1 & 1 0 & 0 & 0 end $ For the set $S_2$, you form the matrix $ egin -2 &-6 & 0 1 & 1 &-2 endsim egin 1 & 3 & 0 1 & 1 &-2 endsim egin 1 & 3 & 0 0 &-2 &-2 endsim egin 1 & 3 & 0 0 & 1 & 1 endsim egin 1 & 0 &-3 0 & 1 & 1 end $

Since we know that elementary row operations do not change row space, we see that $ ewcommand>span(S_1) = span <(1,0,-3),(0,1,1)>= span(S_2).$

This methods works in general. If the non-zero rows in the rrefs are different, that means $span(S_1) e span(S_2)$.

To formulate this more clearly, we know that:

  • The matrices $A$ and $B$ are row equivalent.
  • The matrices $A$ and $B$ have the same row space.
  • The matrices $A$ and $B$ have the same reduced row echelon form.

The beginnings of your strategy were very reasonable. If it had turned out that the collection $S_1$ was linearly independent, then you would have known that the space spanned by $S_1$ has dimension $3$, i.e. is the full space. The space spanned by $S_2$ must have dimension $le 2$, actually exactly $2$. So if $S_1$ had turned out to be a linearly independent set, you could have immediately concluded that the subspaces spanned are different, since they would have different dimension.

Unfortunately, it turned out that the set $S_1$ is linearly dependent, so a simple dimension argument will not solve the problem.

But it is now easy to see that both spaces are $2$-dimensional, they are planes through the origin. If you can show that the two vectors in $S_2$ are each in the span of $S_1$, that will mean the two spaces are the same. If you can show that at least one of the vectors in $S_2$ is not in $S_1$, you will know the spaces are not the same.

Checking this just a matter of looking at some linear equations, but you may be able to eyeball things. For example, the second vector in $S_2$ is the difference between the first two vectors in $S_1$. And the first vector of $S_2$ is (apart from scaling) the sum of the first two vectors in $S_1$. So the spaces are the same.

Since we are in $3$-space, you could alternately, by calculating cross-products, show that the two planes are perpendicular to the same line, and, since they both go through the origin, they are the same plane.

There are many ways to approach this sort of question, based on how much one depends on mechanical approaches vs. intuitive approaches.

Perhaps the most mechanical way of approaching this question is to quite literally Gram-Schmidt these guys to death (as this removes redundant vectors, i.e. linearly dependent vectors, i.e. vectors that are already spanned in the vectors already considered). For example, after one performs G-S on the first set of vectors, we are only left with 2 basis vectors (the third is obliterated). Continuing this, we obliterate the second set of vectors with these two basis vectors. (We note that the second set of vectors, considered separately, are linearly independent - and so also span a space of 2 dimensions).

But what does this accomplish? It's just a mechanical way of explicitly writing each vector in terms of other vectors. So we literally show that anything written in one basis can be written in the other. That's not so exciting.

Suppose we wanted to do it differently still: we can quickly see that the dimension of the space spanned by the first set is 2, and the dimension of the space spanned by the second set is 2. These are so called linear spaces, which is handy. This means that if we find 2 linearly independent vectors contained in both spaces, they are the same. While in one respect, we could just choose the two first vectors and see if they're in the second space, this also means that we could choose any linear combination of the first two vectors as well. So it's slightly more general.

Differently still: each linear subspace is 2 dimensional, i.e. a plane through the origin (it's a linear space, so it contains the origin). You could find the equation of the plane generated by the first two vectors and compare it to the plane generated by the two vectors in the second set. They are the same (multiples of each other). How does one do this? Taking cross products! (if you know them - it's very very simple).

Differently still: after realizing that each subspace is 2 dimensional, throw all 5 vectors in a matrix and row reduce. If 2 are left, they're the same. If there are 3 or more, then they are not the same. Along the same lines, one could (though should not, to be honest) proceed with determinants. Form a matrix whose first and second rows are the vectors from the first set, and whose third row is the first vector of the second set. Form another whose third row is the second vector of the second set. Taking determinants, we will see that both determinants are zero! This means that the 3-dimensional volume of these two parallelopipeds is zero, i.e. that they lie flat! If they lie flat, their sides must be linearly dependent, and since both vectors of the second set are dependent in the first set, they span the same subspace.

Differently still: find a vector not spanned in the first set, find the component orthogonal to the first subspace, and dot this orthogonal component with each vector in the second set. You will get 0 both times, meaning that the two subspaces have the same orthogonal complement, and therefore they are the same. Alternatively, take the cross product of the two vectors in the first set and dot the result with each vector in the second set. You will get 0 again, and this does not bear the burden of finding orthogonal complements in any witty or projective fashion.

Differently still: do it heuristically! By rolling a fair 24-sided die (called a deltoidal icositetrahedron) repeatedly, generate a random set of about 9 different points in 3 space. Find the line of best fit and project onto these two spaces. You will get the same projections! After doing this a couple of times, you can expect this to always work! Afterwords, show that as these 9 points are always on lattice points, at least 1 of the 36 segments joining these 9 points contain a lattice point as well. It will sharpen your skills with pigeonholing ideas.

Differently still, and finally: guess. When I TA'd linear algebra, my students would throw everything they could into matrices and row-reduce them, write a few things about rank and nullity, and solve a linear system of equations (whether asked for or not) on every question. Literally, even this one. When they realized that's not what I asked for, they would write a few illegible lines of work, draw a big $Longrightarrow$, and say something like "Clearly, they do not span the same space" or "Obviously, they span the same space." For some other TAs, they'd give partial credit. It at least gave me a laugh.

Seriously though, if you would like to explain any of these further, let me know. They all really do work. And yes, it was quite a yarn. But I've been gone for a couple of weeks, and I had to say hello somehow!


1 Answer 1

To assist with your problem, I will describe the general approach, when we don't necessarily know what we're looking for. As you are aware, finding the Killing vectors $X^mu$ requires solving,

$ abla_mu X_ u + abla_ u X_mu = 0$

which is an over-determined system of differential equations. As you know, these Killing vectors are precisely those for which $mathcal L_X g = 0$ and are the infinitesimal generators of isometries of the manifold being described. If you split them into a linearly independent set, $$ you can work out the Lie algebra they generate and sometimes may be able to identify what they are physically.

Alternatively, one can explicitly find the finite form of the transformations they generate by solving the equation for the integral curves of the fields, that is,

where $x^mu(lambda)$ parametrises the integral curve, that is, you can think of it as an embedding function and $X^mu(x)$ is the Killing field, seen as a function of the components $x^mu$.

Choosing the right coordinates often makes the calculation simpler it is certainly desirable to have the metric be diagonal, and well-behaved at as many points as possible.


An embedding is a matrix in which each column is the vector that corresponds to an item in your vocabulary. To get the dense vector for a single vocabulary item, you retrieve the column corresponding to that item.

But how would you translate a sparse bag of words vector? To get the dense vector for a sparse vector representing multiple vocabulary items (all the words in a sentence or paragraph, for example), you could retrieve the embedding for each individual item and then add them together.

If the sparse vector contains counts of the vocabulary items, you could multiply each embedding by the count of its corresponding item before adding it to the sum.

These operations may look familiar.

Embedding lookup as matrix multiplication

The lookup, multiplication, and addition procedure we've just described is equivalent to matrix multiplication. Given a 1 X N sparse representation S and an N X M embedding table E, the matrix multiplication S X E gives you the 1 X M dense vector.

But how do you get E in the first place? We'll take a look at how to obtain embeddings in the next section.

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