What is the unit vector that is orthogonal to the plane containing $(2i + 3j – 7k)$ and $(3i - j - 2k)$ ?

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Answer 1 · 2016-11-30T16:36:09

The answer is $=1/sqrt579*〈-13,-17,-11〉$

To calculate a vector perpendicular to two other vectors, you have to calculate the cross product

Let $vecu=〈2,3,-7〉$ and $vecv=〈3,-1,-2〉$

The cross product is given by the determinant

$| (i,j,k), (u_1,u_2,u_3), (v_1,v_2,v_3) |$

$vecw=| (i,j,k), (2,3,-7), (3,-1,-2) |$

$=i(-6-7)-j(-4+21)+k(-2-9)$

$=i(-13)+j(-17)+k(-11)$

$=〈-13,-17,-11〉$

To verify that $vecw$ is perpendicular to $vecu$ and $vecv$

We do a dot product.

$vecw.vecu=〈-13,-17,-11〉.〈2,3,-7〉=-26--51+77=0$

$vecw.vecv=〈-13,-17,-11〉.〈3,-1,-2〉=-39+17+22=0$

As the dot products $=0$ , $vecw$ is perpendicular to $vecu$ and $vecv$

To calculate the unit vector, we divide by the modulus

$hatw=vecw/(∥vecw∥)=1/sqrt579*〈-13,-17,-11〉$

What is the unit vector that is orthogonal to the plane containing (2i + 3j – 7k) (2i + 3j – 7k) and (3i - j - 2k) (3i - j - 2k) ?