Question

Find a unit vector perpendicular to each of the vectors $2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-2\overrightarrow{b}$, where $\overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$.

Answer 1

Given $\overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$

Thus $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 1& 1& 1\end{array}\right|=\hat{i}(2+1)-\hat{j}(1+1)+\hat{k}(1-2)=3\hat{i}-2\hat{j}-\hat{k}$

Now vector perpendicular to $2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-2\overrightarrow{b}$ is given by,

$(2\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-2\overrightarrow{b})=2(\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{b}\times \overrightarrow{a})-(4\overrightarrow{a}\times b)-2(\overrightarrow{b}\times \overrightarrow{b})$

$=2\left(0\right)-\overrightarrow{a}\times \overrightarrow{b}-4(\overrightarrow{a}\times \overrightarrow{b})-2\left(0\right)=-5\overrightarrow{a}\times \overrightarrow{b}$

$=-5(3\hat{i}-2\hat{j}-\hat{k})$

So required unit vector is $=\pm \frac{3\hat{i}-2\hat{j}-\hat{k}}{\sqrt{3^2+2^2+1^2}}=\pm \frac{3\hat{i}-2\hat{j}-\hat{k}}{\sqrt{14}}$