Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\mu \hat{j}+u \ha

Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\mu \hat{j}+u \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} 	imes \hat{d}=\vec{b} 	imes \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \vec{c}|^{2}$ is equal to $\_\_\_\_$

Professor PC Thomas Classes · Answer

(5) : We have, $\vec{a} 	imes \hat{d}-\vec{b} 	imes \hat{d}=0$\ $\Rightarrow(\vec{a}-\vec{b}) 	imes \hat{d}=0$\ $	herefore \quad \hat{d}=t(\vec{a}-\vec{b})$, for any scalar $t$.\ $\Rightarrow \hat{d}=t(-2 \hat{i}-\hat{j}+2 \hat{k})$\ Since, $|\hat{d}|=1$\ $	herefore|t|=\frac{1}{3}$\ Now, $\vec{c} \cdot \vec{a}=0[\because \vec{c}$ is perpendicular to $\vec{a}]$\ $\Rightarrow \lambda+\mu=0 \Rightarrow \mu=-\lambda$\ $	herefore \quad \vec{c}=\lambda(\hat{j}-\hat{k}) \Rightarrow|\vec{c}|^{2}=2 \lambda^{2}$\ Also, $\vec{c} \cdot \hat{d}=1$\ $\Rightarrow \lambda(\hat{j}-\hat{k}) \cdot t(-2 \hat{i}-\hat{j}+2 \hat{k})=1$\ $\Rightarrow \lambda t=\frac{-1}{3} \Rightarrow \lambda^{2}=1 \quad\left[\because|t|=\frac{1}{3}ight]$\ $	herefore|3 \lambda \hat{d}+\mu \vec{c}|^{2}=9 \lambda^{2}|\hat{d}|^{2}+\mu^{2}|\vec{c}|^{2}+6 \lambda \mu(\hat{d} \cdot \vec{c})$ $$ =3 \lambda^{2}+2 \lambda^{4}=5 $$