Find any vector w that is perpendicular to both u = 3j + 4k and v = 2i.

Question

i, j, k are basis vectors.

Textbook answer: (0,4,-3)

My attempt: used any scale factor variation, like 4j-3k or -8j+6k to get w=8j-6k

How do i solve this the longer way?

Answer 1

You want the cross-product of <0,3,4> and <2,0,0>
You must have a method of finding that, by the method I use
I got <0,8,-6> which is "lowest terms" would be <0,4,-3>

Here is Sal showing you a method which is easy to follow. I basically
use the same method just lined up in a small variation of that.
https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/dot-cross-products/v/linear-algebra-cross-product-introduction

Answer 2

I guess but I like the way I learned better:

A x B =
determinant
| i j k |
| 0 3 4|
| 2 0 0 |
= (3*0-0*4) i + (4*2-0*0) j + (0*0-3*2) k
= 0 i + 8 j - 6 k

Answer 3

That's the same way I do it, just a variation of what Sal did in the video.

Answer 4

think of it another way.
Any vector (0,y,z) will be perpendicular to v.
So, now you just need a line in y-z space perpendicular to (3,4).
In terms of slopes, (3,4) has slope 4/3
So, a vector with slope -3/4 will be perpendicular.

So, (0,4,-3) and (0,3,-4) will be perpendicular to u and v as given.