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

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Answer 1 · 2016-09-15T04:37:40

$+-1/sqrt 14(3, 1, -2)$ , for opposite directions.

If vector $u=(u_1, u_2, u_3) and v=(v_1, v_2, v_3)$ , then

vectors $+-uXv=+-(u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1)$

are perpendicular to both $u and v$ , in the opposite directions.

Here, $u(0, 2, 1) and v=(1, -1, 1)$ . So,

$+-uXv$

$=+-((2)(1)-(1)(-1), (1)(1)-(0)((1), (0)(-1)-((2)(1))$

$=+-(3, 1, -2)$ .

divide by the modulus $|(3, 1, -2)|=sqrt(3^2+(1)^2+(-2)^2)=sqrt 14$ ,