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. Then the value of $\lambda$ is

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{b}=2\hat{i}+4\hat{j}-5\hat{k}$
$\overrightarrow{c}=\lambda \hat{i}+2\hat{j}+3\hat{k}$
So, $\overrightarrow{b}+\overrightarrow{c}=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$
Unit vector along$\overrightarrow{b}+\overrightarrow{c}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+36+4}}$
Given that dot product of $\overrightarrow{a}$ with the unit vector along the $\overrightarrow{b}+\overrightarrow{c}$ is equal to $1$.
So,$\left(\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+40}}\right)\cdot (\hat{i}+\hat{j}+\hat{k})=1$
$\Rightarrow \frac{(2+\lambda )+6-2}{\sqrt{(2+\lambda {)}^2+40}}=1$
$\Rightarrow (2+\lambda )+4=\sqrt{(2+\lambda {)}^2+40}$
Squaring both sides,
$\Rightarrow (2+\lambda {)}^2+16+8(2+\lambda )=(2+\lambda {)}^2+40$
$\Rightarrow 32+8\lambda =40$
$\Rightarrow 8\lambda =8$
$\Rightarrow \lambda =1$