Question

Determine a unit vector perpendicular to both a=2i+j+k and b=i-j+2k

Answer 1

Answer: $\frac{1}{9 } (\hat{i}-\hat{j}-\hat{k})$

Step-by-step explanation:

Let $\vec{c}$ be the perpendicular vector to both vectors $\vec{a}$ and $\vec{b}$

Thus, $\vec{c}=\vec{a}\times \vec{b}$

$\vec{c} =  \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2\end{vmatrix}$

$= \hat{i}(2+1)-\hat{j}(4-1)+\hat{k}(-2-1)$

$= 3\hat{i} -3\hat{j} - 3\hat{k}$

Thus, the unit vector of $\vec{c}$ is,

$\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3\hat{i}-3\hat{j}-3\hat{k}}{\sqrt{3^2+3^2+3^2} } = \frac{1}{\sqrt{729} } (3\hat{i}-3\hat{j}-3\hat{k})= \frac{1}{27 } (3\hat{i}-3\hat{j}-3\hat{k}) = \frac{1}{9 } (\hat{i}-\hat{j}-\hat{k})$

Answer 2

Here's your answer

Hope it's helpful