Mathematics - Vector Question with Solution | TestHub

Let $\vec{d}=\vec{b}+\lambda \vec{c}$ $$\begin{array}{rl}& \therefore \vec{d}=\hat{i}+2\hat{j}-\hat{k}+\lambda (\hat{i}+\hat{j}-2\hat{k})\\ & =(1+\lambda )\hat{i}+(2+\lambda )\hat{j}-(1+2\lambda )\hat{k}\end{array}$$ If $\theta$ be the angle between $\vec{d}$ and $\vec{a}$, then projection of $\vec{d}$ or $(\vec{b}+\lambda \vec{c})$ on $\vec{a}$ $$\begin{array}{rl}& =|\vec{d}|\cos \theta \\ & =|\vec{d}|\left(\frac{\vec{d}\cdot \vec{a}}{|\vec{d}||\vec{a}|}\right)=\frac{\vec{d}\cdot \vec{a}}{|\vec{a}|}\\ & =\frac{2(\lambda +1)-(\lambda +2)-(2\lambda +1)}{\sqrt{4+1+1}}\\ & =\frac{-\lambda -1}{\sqrt{6}}\end{array}$$ But projection of $\vec{d}$ on $\vec{a}=\sqrt{\frac{2}{3}}$ $$\therefore -\frac{\lambda +1}{\sqrt{6}}=\sqrt{\frac{2}{3}}\Rightarrow \frac{{\lambda }^2+2\lambda +1}{6}=\frac{2}{3}$$ $$\begin{array}{rl}& \Rightarrow {\lambda }^2+2\lambda -3=0\Rightarrow {\lambda }^2+3\lambda -\lambda -3=0\\ & \Rightarrow \lambda (\lambda +3)-1(\lambda +3)=0,\Rightarrow \lambda =1,-3\end{array}$$ when $\lambda =1$, then $\vec{b}+\lambda \vec{c}=2\hat{i}+3\hat{j}-3\hat{k}$ when $\lambda =-3$, then $\vec{b}+\lambda \vec{c}=-2\hat{i}-\hat{j}+5\hat{k}$