Question

Let vectors $a=6i-3j-k$ and $b=2i+3j+k$, find a unit vector which is perpendicular to the vectors $a+b$ and $a-b$.

Answer 1

$\begin{array}{c}\Rightarrow \left(a+b\right)=8i+0\hat{j}+0\hat{k}\left[a\perp b,a.b=0\right]\\ \Rightarrow \left(a-b\right)=4i-6\hat{j}-2\hat{k}\\ Letunitvectorni+yj+2\hat{k}\\ \Rightarrow \left(xi+yj+z\hat{k}\right).8i=8k=0\\ \Rightarrow \left(4i-6\hat{j}-2\hat{k}\right).\left(xi+yj+z\hat{k}\right)=0\\ \Rightarrow 4x-6y-2z=0\\ 3y+z=0\to \left(i\right)\\ \Rightarrow x^2+y^2+z^2=1\\ \Rightarrow y^2+z^2=1\\ fromed\left(i\right),{\left(3y\right)}^2=y^2=1,10y^2=1,y=\pm \frac{1}{\sqrt{10}}\\ Puty=\pm \frac{1}{\sqrt{10}}inequation\left(i\right),z=\pm \frac{3}{\sqrt{10}}\\ Then,unitvectoris\left(0i+\frac{1}{\sqrt{10}}y-\frac{3}{\sqrt{10}}\hat{k}\right)or\left(\frac{-\hat{j}}{\sqrt{10}}+\frac{3}{\sqrt{10}}\hat{k}\right)\end{array}$