Question

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda \hat{i}+2\hat{j}+3\hat{k}$ is equal to one. Find the value of $\lambda$.

Answer 1

Let $\overrightarrow{a}=2\hat{i}+4\hat{j}-5\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+2\hat{j}+3\hat{k}$
Then,
$\overrightarrow{a}+\overrightarrow{b}=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$
Unit vector along $\overrightarrow{a}+\overrightarrow{b}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+6^2+(-2{)}^2}}$
Given, $(\hat{i}+\hat{j}+\hat{k}).\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+6^2+(-2{)}^2}}=1$
$\Rightarrow 1\times (2+\lambda )+1\times 6+1\times (-2)=\sqrt{(2+\lambda {)}^2+40}$
$\Rightarrow \lambda +6=\sqrt{(2+\lambda {)}^2+40}$
$\Rightarrow {\lambda }^2+12\lambda +36=4+4\lambda +{\lambda }^2+40$
$\Rightarrow 12\lambda +36=4\lambda +44\Rightarrow \lambda =1$