Question

The point of intersection of lines
$L_1:(1+\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+3\lambda )\hat{k},\lambda \in R$
$L_2:(3+2\mu )\hat{i}+(4+2\mu )\hat{j}+(1-2\mu )\hat{k},\mu \in R$ is

• A. $(-1,2,3)$
• B. $(1,2,3)$
• C. $(-2,3,4)$
• D. $(3,4,-1)$

Answer 1

The correct option is B $(1,2,3)$

Given lines

$L_1:(1+\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+3\lambda )\hat{k}$

$L_2:(3+2\mu )\hat{i}+(4+2\mu )\hat{j}+(1-2\mu )\hat{k}$

$L_2:3\hat{i}+4\hat{j}+\hat{k}+\mu (2\hat{i}+2\hat{j}-2\hat{k})$

$L_1:\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}+2\hat{j}+3\hat{k})$

Given line $L_1$ in cartesian form

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}=\lambda$

Hence the point of intersection of line $L_1$ and $L_2$

$I(\lambda +1,2\lambda +2,3\lambda +3)$

Line $L_2$ in cartesian form

$\frac{x-3}{2}=\frac{y-4}{2}=\frac{z-1}{-2}$

The point of intersection also lie on line $L_2$

Hence $\frac{\lambda +1-3}{2}=\frac{2\lambda +2-4}{2}=\frac{3\lambda +3-1}{-2}$

$\frac{\lambda +1-3}{2}=\frac{2\lambda +2-4}{2}$

$\lambda -2=2\lambda -2$

$\lambda =0$

Putiing $\lambda$ in point I

$I(1,2,3)$