vectors

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?

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  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

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  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

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

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  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.

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