Question

If V=⟨3,2⟩ and W=⟨-1,-3⟩, find ll V x W ll
Can someone please explain how the cross product thing works?

Answer 1

If Z = <a, b> is the cross product of V and W, then Z is perpendicular to both V and W

So this must mean

Z dot V = 0

Z dot W = 0

Answer 2

<a, b> dot <3, 2> = 3a + 2b = 0

<a, b> dot <-1, -3> = -a - 3b = 0

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So you have these two equations

3a+2b = 0
-a - 3b = 0

which you can use to solve for a and b

Answer 3

The double vertical bars around V x W tells you to find the magnitude/length of this vector

Answer 4

oh wait...the cross product is only possible in R3 lol, my bad

Answer 5

@jim_thompson5910 Cross product only works well in 3 dimensions.

Answer 6

yeah just remembered lol

Answer 7

well you can fix that by just adding on a z-coordinate of 1 or 0 or something

Answer 8

Haha yeah I saw that in my notes and that's why I got confused.. So it is still possible to solve then?

Answer 9

V = <3, 2> could be thought of as V = <3, 2, 0>

W = <-1,-3> could be thought of as W = <-1, -3, 0>

since you can think of the xy plane as the plane where z = 0 (in 3 space)

Answer 10

hmm maybe it might be better to make the third coordinate 1...not sure

Answer 11

\[
\|\mathbf v \times \mathbf w\| = \|\mathbf v\|\|\mathbf w\|\sin(\theta)
\]

Answer 12

\[
\sin(\theta) = \sqrt{1-\cos^2(\theta)}
\]\[
\cos(\theta) = \frac{\|\mathbf v\cdot \mathbf w\|}{\|\mathbf v\|\|\mathbf w\|}
\]

Answer 13

V = <3, 2> could be thought of as V = <3, 2, 1>

W = <-1,-3> could be thought of as W = <-1, -3, 1>

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If Z = <a, b, c> then

Z dot V = <a, b, c> dot <3, 2, 1> = 3a + 2b + c = 0

Z dot W = <a, b, c> dot <-1, -3, 1> = -a - 3b + c = 0

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So you have these two equations

3a + 2b + c = 0
-a - 3b + c = 0

it's still possible to solve this system, but it will have more than one solution

Answer 14

\[
\|\mathbf v \times \mathbf w\| = \sqrt{(\|\mathbf v\|\|\mathbf w\|)^2-(\|\mathbf v\cdot \mathbf w\|)^2}
\]This is my best generalization of the magnitude of the cross product

What do you need it anyway?

Answer 15

@wio Ya that equation makes the most sense. Oh and it's just a homework problem

Answer 16

Thank you!!

Answer 17

Basically the subtraction of the dot product is trying to work against any parallelness.