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In the question number 20, a unit vector perpendicular to the direction of `vecA and vecB ` is
A. `(-2hati-hatj-hatk)/(sqrt(6))`
B. `(2hati+hatj+hatk)/(sqrt(6))`
C. `(2hati-hatj-hatk)/(sqrt(6))`
D. `(2hati-hatj+hatk)/(sqrt(6))`

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Correct Answer - A
Unit vector perpendicular to the direction of `vecA and vecB` is `hatn = ((vecA xx vecB))/(|vecAxx vecB|)`
`vecA xx vecB = |{:(hati,,hatj,,hatk),(-2,,3,,1),(1,,2,,-4):}|`
` = hati (-12-2) + hatj(1-8) + hatk(-4-3) = -14hati - 7hatj - 7hatk`
`|vecAxxvecB| = sqrt((1-14)^(2) + (-7)^(2) +(-7)^(2)) = 7sqrt(6)`
`therefore hatn = (-14hati - 7hatj -7hatk)/(7sqrt(6)) = (-2hati-hatj -hatk)/(sqrt(6))`

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