Question

If $\vec{a}=\hat{i}+\lambda \hat{j}+2\hat{k}$ and $\vec{b}=\mu \hat{i}+\lambda \hat{j}+2\hat{k}$ are orthogonal and if $|\vec{a}|=|\vec{b}|$, then $(\lambda ,\mu )$ =

• A. $\left(\frac{1}{4},-\frac{9}{4}\right)$
• B. $\left(-\frac{1}{4},\frac{9}{4}\right)$
• C. $\left(\frac{7}{4},\frac{1}{4}\right)$
• D. $\left(\frac{1}{4},\frac{7}{4}\right)$

Answer 1

Answer: (D)

The vectors $\vec{a}=\hat{i}+\lambda \hat{j}+2\hat{k}$ and $\vec{b}=\mu \hat{i}+\lambda \hat{j}+2\hat{k}$ are orthogonal .
$\therefore \vec{a}.\vec{b}=0$
$\Rightarrow \mu +\lambda -2=0\Rightarrow \mu +\lambda =2$ ....(i)
Since , $|\vec{a}|=|\vec{b}|$
$\therefore \sqrt{1+{\lambda }^2+4}=\sqrt{{\mu }^2+1+1}\Rightarrow {\lambda }^2+5={\mu }^2+2$
$\Rightarrow {\lambda }^2+5={\left(2-\lambda \right)}^2+2$ [using (i)]
$\Rightarrow {\lambda }^2+5=4+{\lambda }^2-4\lambda +2$
$\Rightarrow \lambda =\frac{1}{4}$
$\therefore \mu =2-\frac{1}{4}=\frac{7}{4}$ Hence $\left(\lambda ,\mu \right)=\left(\frac{1}{4},\frac{7}{4}\right)$