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Q. 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 $(\lambda, \mu)$ =

COMEDKCOMEDK 2008Vector Algebra

Solution:

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)$