Asked by annonymous
                Find a unit vector perpendicular to a=(4,-3,1) and b=(2,3,-1)
What I did:
a x b = ((-3)(-1)-(3)(1), (1)(2)-(-1)(4), (4)(3) - (2)(-3))
=(0,6,18)
Answer is (0, 1/square root 10, 3/square root 10)
            
        What I did:
a x b = ((-3)(-1)-(3)(1), (1)(2)-(-1)(4), (4)(3) - (2)(-3))
=(0,6,18)
Answer is (0, 1/square root 10, 3/square root 10)
Answers
                    Answered by
            Reiny
            
    your initial vector is correct 
magnitude of (0,6,18) = √(0+36+324) = √360
= 6√10
so I see that your unit vector is also correct
Perhaps you are looking at some answer key and the answer there looks different.
They might have rationalized the denominator
1/√10
= 1/√10 *√10/√10
= √10/10
3/√10 = 3√10/10
alternate form of your unit vector:
(0 , √10/10 , 3√10/10 )
    
magnitude of (0,6,18) = √(0+36+324) = √360
= 6√10
so I see that your unit vector is also correct
Perhaps you are looking at some answer key and the answer there looks different.
They might have rationalized the denominator
1/√10
= 1/√10 *√10/√10
= √10/10
3/√10 = 3√10/10
alternate form of your unit vector:
(0 , √10/10 , 3√10/10 )
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