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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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

$\pm \frac{1}{\sqrt{14}} (3, 1, - 2)$ $+-1/sqrt 14(3, 1, -2)$ , for opposite directions.

Explanation:

If vector $u = (u_{1}, u_{2}, u_{3}) and v = (v_{1}, v_{2}, v_{3})$ $u=(u_1, u_2, u_3) and v=(v_1, v_2, v_3)$ , then

vectors $\pm u X v = \pm (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1})$ $+-uXv=+-(u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1)$

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

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

$\pm u X v$ $+-uXv$

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

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

For unit vectors,

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

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

1 Answer

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