Question

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

Answer 1

Given:

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=2\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{c}=\lambda \hat{i}+2\hat{j}+3\hat{k}$

So, $\overrightarrow{b}+\overrightarrow{c}=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$

Unit vector along $\overrightarrow{b}+\overrightarrow{c}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+36+4}}$

$=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+40}}$

Given that dot product of $\overrightarrow{a}$ with the unit vector of $\overrightarrow{b}+\overrightarrow{c}$ is equal to $1.$

So, apply given condition,

$\frac{(2+\lambda )+6-2}{\sqrt{(2+\lambda {)}^2+40}}=1$

$\Rightarrow 2+\lambda +4=\sqrt{(2+\lambda {)}^2+40}$

Squaring, we get

$36+{\lambda }^2+12\lambda =4+{\lambda }^2+4\lambda +40$

$\Rightarrow 8\lambda =8$

$\Rightarrow \lambda =1$