How do you find a unit vector perpendicular to two vectors that is perpendicular to both the vectors u = (0, 2, 1) and v = (1, -1, 1)?

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Answer 1 · 2016-07-22T17:14:15

Reqd. vector $= (\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, - \frac{2}{\sqrt{14}})$ $=(3/sqrt14,1/sqrt14,-2/sqrt14)$ .

A well-known Property of the Vector Product will be useful in this case.

Given two vectors $\to x and \to y$ $vecx and vecy$ , we know that, $\to x$ $vecx$ x $\to y$ $vecy$

is a vector that is $⊥$ $bot$ to both $\to x & \to y$ $vecx & vecy$

Therefore, taking $\to u \times \to v = \to w,$ $vecu xx vecv = vec w,$ say, we get,

$vecw=|(hati, hatj, hatk), (0,2,1), (1,-1,1)|$

$=3hati+hatj-2hatk=(3,1,-2)$

Now reqd. unit vector, i.e., $hatw$ is given by, $vecw/||vecw||$ ,

where, $||vecw||=sqrt(3^2+1^2+(-2)^2)=sqrt14$

Hence, reqd. vector $hatw=(3/sqrt14,1/sqrt14,-2/sqrt14)$ .