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

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?

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

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

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

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.

To find a vector w that is perpendicular to both u = 3j + 4k and v = 2i, you can use the method of taking the cross product of the two vectors.

The cross product of two vectors gives you a new vector that is perpendicular to both of the original vectors. Here's how you can use this method to solve the problem:

1. Write the vectors u and v as column vectors:
u = [0, 3, 4]
v = [2, 0, 0]

2. To find the cross product of u and v, take the determinant of the following matrix:
| i j k |
| 0 3 4 |
| 2 0 0 |

The determinant can be calculated using the formula:
determinant = i(3*0 - 4*0) - j(0*0 - 4*2) + k(0*0 - 3*2)

3. Simplify the determinant:
determinant = 0 - 0 + (0 - 6)k = -6k

4. The resulting vector w, which is perpendicular to both u and v, is the vector obtained by setting the coefficients of i, j, and k in the determinant equal to the components of w. Since the coefficient of k is -6, we have:
w = [0, 0, -6]

Therefore, the longer way to solve this problem gives us the vector w = [0, 0, -6]. However, note that the textbook answer, w = [0, 4, -3], is also a correct answer. Vectors can have multiples that are also perpendicular to the original vectors. In this case, both [0, 0, -6] and [0, 4, -3] are multiples of each other, and they are both perpendicular to u and v.