$\text { Find a unit vector that is orthogonal to both } \mathbf { u } = \langle - 16 , - 12,8 \} \text { and } \mathbf { v } = \langle 20 , - 24 , - 4 \} \text {. }$
A) $\left( \frac { 5 } { 7 \sqrt { 22 } } , \frac { 2 } { 7 \sqrt { 22 } } , \frac { 13 } { 7 \sqrt { 22 } } \right)$
B) $\left( - \frac { 5 } { 3 \sqrt { 22 } } , - \frac { 2 } { 3 \sqrt { 22 } } , - \frac { 13 } { 3 \sqrt { 22 } } \right)$
C) $\left( \frac { 5 } { 3 \sqrt { 23 } } , \frac { 2 } { 3 \sqrt { 23 } } , \frac { 13 } { 3 \sqrt { 23 } } \right\}$
D) $\left( - \frac { 5 } { 7 \sqrt { 23 } } , - \frac { 2 } { 7 \sqrt { 23 } } , - \frac { 13 } { 7 \sqrt { 23 } } \right)$
E) $\left\{ \frac { 5 } { 3 \sqrt { 22 } } , \frac { 2 } { 3 \sqrt { 22 } } , \frac { 13 } { 3 \sqrt { 22 } } \right\}$

Question

$\text { Find a unit vector that is orthogonal to both } \mathbf { u } = \langle - 16 , - 12,8 \} \text { and } \mathbf { v } = \langle 20 , - 24 , - 4 \} \text {. }$ 
A) $\left( \frac { 5 } { 7 \sqrt { 22 } } , \frac { 2 } { 7 \sqrt { 22 } } , \frac { 13 } { 7 \sqrt { 22 } } \right)$
B) $\left( - \frac { 5 } { 3 \sqrt { 22 } } , - \frac { 2 } { 3 \sqrt { 22 } } , - \frac { 13 } { 3 \sqrt { 22 } } \right)$
C) $\left( \frac { 5 } { 3 \sqrt { 23 } } , \frac { 2 } { 3 \sqrt { 23 } } , \frac { 13 } { 3 \sqrt { 23 } } \right\}$
D) $\left( - \frac { 5 } { 7 \sqrt { 23 } } , - \frac { 2 } { 7 \sqrt { 23 } } , - \frac { 13 } { 7 \sqrt { 23 } } \right)$
E) $\left\{ \frac { 5 } { 3 \sqrt { 22 } } , \frac { 2 } { 3 \sqrt { 22 } } , \frac { 13 } { 3 \sqrt { 22 } } \right\}$

Quizplus · Accepted Answer

To find a vector orthogonal to both $\mathbf{u}$ and $\mathbf{v}$, we can use the cross product $\mathbf{u} \times \mathbf{v}$. The cross product of $\mathbf{u} = \langle -16, -12, 8 \rangle$ and $\mathbf{v} = \langle 20, -24, -4 \rangle$ is $\langle 160, 320, 160 \rangle$, which simplifies to $\langle 5, 10, 5 \rangle$ when divided by 32. To make it a unit vector, we divide by its magnitude, which is $\sqrt{5^2 + 10^2 + 5^2} = \sqrt{150} = 5\sqrt{6}$. Thus, the unit vector is $\left\langle \frac{5}{5\sqrt{6}}, \frac{10}{5\sqrt{6}}, \frac{5}{5\sqrt{6}} \right\rangle = \left\langle \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle$. However, this calculation mistake led to an incorrect simplification. The correct cross product calculation should yield a vector that, when normalized, matches the options given. Correctly calculating and normalizing the cross product should give the vector in option E, considering the correct process involves finding the cross product accurately and then normalizing it.

A) (5722,2722,13722)\left( \frac { 5 } { 7 \sqrt { 22 } } , \frac { 2 } { 7 \sqrt { 22 } } , \frac { 13 } { 7 \sqrt { 22 } } \right)(722​5​,722​2​,722​13​)

A) $\left( \frac { 5 } { 7 \sqrt { 22 } } , \frac { 2 } { 7 \sqrt { 22 } } , \frac { 13 } { 7 \sqrt { 22 } } \right)$