Question

If the projection of the vector $\hat{i}+2\hat{j}+\hat{k}$ on the sum of the two vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $-\lambda \hat{i}+2\hat{j}+3\hat{k}$ is $1,$ then $\lambda$ is equal to

Answer 1

$\overrightarrow{v_1}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{v_2}=2\hat{i}+4\hat{j}-5\hat{k},\overrightarrow{v_3}=-\lambda \hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{v_2}+\overrightarrow{v_3}=(2-\lambda )\hat{i}+6\hat{j}-2\hat{k}=\overrightarrow{v_4}$

Projection of $\overrightarrow{v_1}$ on $\overrightarrow{v_4}=\overrightarrow{v_1}\cdot \frac{\overrightarrow{v_4}}{|\overrightarrow{v_4}|}$
$\Rightarrow \frac{1\times (2-\lambda )+2\times 6+1\times (-2)}{\sqrt{(2-\lambda {)}^2+6^2+(-2{)}^2}}=1$
$\Rightarrow (12-\lambda {)}^2=(2-\lambda {)}^2+40$
On solving, we get
$\lambda =5$