Mathematics - Vector Question with Solution | TestHub

Let $x\hat{i}+y\hat{j}+z\hat{k}$ be the required unit vector. Since $\hat{a}$ is perpendicular to $(2\hat{i}-\hat{j}+2\hat{k})$. $$\therefore 2x-y+2z=0$$ Since vector $x\hat{i}+y\hat{j}+z\hat{k}$ is coplanar with the vector $\hat{i}+\hat{j}-\hat{k}$ and $2\hat{i}+2\hat{j}-\hat{k}$. $$\begin{array}{rl}\therefore x\hat{i}& +y\hat{j}+z\hat{k}\\ & =p(\hat{i}+\hat{j}-\hat{k})+q(2\hat{i}+2\hat{j}-\hat{k}),\end{array}$$ where $p$ and $q$ are some scalars. $$\begin{array}{rl}& \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}\\ & =(p+2q)\hat{i}+(p+2q)\hat{j}-(p+q)\hat{k}\\ & \Rightarrow x=p+2q,y=p+2q,z=-p-q\end{array}$$ Now from equation (i), $$\begin{array}{rl}& 2p+4q-p-2q-2p-2q=0\\ \Rightarrow & -p=0\Rightarrow p=0\\ \therefore & x=2q,y=2q,z=-q\end{array}$$ Since vector $x\hat{i}+y\hat{j}+z\hat{k}$ is a unit vector, therefore $$\begin{array}{rl}& |x\hat{i}+y\hat{j}+z\hat{k}|=1\\ \Rightarrow & \sqrt{x^2+y^2+z^2}=1\\ \Rightarrow & x^2+y^2+z^2=1\\ \Rightarrow & 4q^2+4q^2+q^2=1\end{array}$$ $$\Rightarrow 9q^2=1\Rightarrow q=\pm \frac{1}{3}$$ When $q=\frac{1}{3}$, then $x=\frac{2}{3},y=\frac{2}{3}$, $$z=-\frac{1}{3}$$ When $q=-\frac{1}{3}$, then $x=-\frac{2}{3},y=-\frac{2}{3}$, $$z=\frac{1}{3}$$ Here required unit vector is $\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}$ or $-\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}$.