Question

Find a unit vector perpendicular to each of the vectors a + b and a - b, where $a=3\hat{i}+2\hat{j}+2\hat{k}$ and $b=\hat{i}+2\hat{j}-2\hat{k}$

Answer 1

Given that $a=3\hat{i}+2\hat{j}+2\hat{k}$ and $b=\hat{i}+2\hat{j}-2\hat{k}$
$a+b=(3\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k})=4\hat{i}+4\hat{j}+0\hat{k}$
and $a-b=(3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})=2\hat{i}+4\hat{k}$
Now, $(a+b)\times (a-b)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|$
$=\hat{i}(16-0)-\hat{j}(16-0)+\hat{k}(0-8)=16\hat{i}-16\hat{j}-8\hat{k}\Rightarrow |(a+b)\times (a-b)|=\sqrt{(16{)}^2+(-16{)}^2+(-8{)}^2}=\sqrt{256+256+64}=\sqrt{576}=24$
$\therefore$ A unit vector, perpendicular to both (a + b) and (a - b) is
$\pm \frac{(a+b)\times (a-b)}{|(a+b)\times (a-b)|}=\pm \frac{16\hat{i}-16\hat{j}-8\hat{k}}{24}=\pm \frac{8(2\hat{i}-2\hat{j}-1\hat{k})}{24}=\pm \frac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})$
$\therefore$ Required vector is either $\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}$ or $-\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}$