I thought this was straightforward. I found cross product of both vectors (I ass

Question

I thought this was straightforward. I found cross product of both vectors (I assume they are in format already that i can use.

I arrived at 9i - 12j +12k. I then standardized that by dividing by magnitude of that which was sqrt of 81+144+144, or sqrt of 369.

I don't know how I get my answer (i.e. the perpendicular vector found by cross product) to start with a '1' in the x component part of the vector description.

Please show each and every step you did in solving this problem. No points for work not shown.

Thanks.

Explanation / Answer

Cross product of a vector is a vector which is perpendicular to both the vectors you are talking cross cross product off. The cross product of given two vectors is :

9i - 12j +12k

Now, what you need is direction vector unrelated to its magnitude.

So divide the above answer with 9, you get,

i - 4/3 j + 4/3 k

So your answer is <1,-4/3,4/3>

What you were trying to find was a unit vector i.e. a vector whose magnitude is one , in the direction perpendicular to given two vectors .

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