Question

If $L_1$ is a line through the point $5\hat{i}+8\hat{j}+11\hat{k}$ and parallel to the vector $2\hat{i}+3\hat{j}+4\hat{k}$ and $L_2$ is a line through the point $4\hat{i}+6\hat{j}+8\hat{k}$ and parallel to the vector $3\hat{i}+4\hat{j}+5\hat{k}$, then the point of intersection of $L_1$ and $L_2$ is

• A. $\hat{i}+\hat{j}+\hat{k}$
• B. $\hat{i}+2\hat{j}+3\hat{k}$
• C. $2\hat{i}+3\hat{j}+\hat{k}$
• D. $\hat{i}-2\hat{j}+2\hat{k}$

Answer 1

Answer: (B)

Line $L_1$ is passing through $5\hat{i}+8\hat{j}+11\hat{k}$ and parallel to the vector $2\hat{i}+3\hat{j}+4\hat{k}$.
$\therefore L_1=r=5\hat{i}+8\hat{j}+11\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$
line $L_2$ is passing through $4\hat{i}+6\hat{j}+8\hat{k}$ and parallel to the vector $3\hat{i}+4\hat{j}+5\hat{k}$
$\therefore L_2=r=4\hat{i}+6\hat{j}+8\hat{k}+\mu (3\hat{i}+4\hat{j}+5\hat{k})$
$L_1$ and $L_2$ are intersecting.
$\therefore 5\hat{i}+8\hat{j}+11\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})=4\hat{i}+6\hat{j}$
$+8\hat{k}+\mu (3\hat{i}+4\hat{j}+5\hat{k})$
$\Rightarrow \hat{i}+2\hat{j}+3\hat{k}=(3\mu -2\lambda )\hat{i}+(4\mu -3\lambda )\hat{j}+(5\mu -4\lambda )\hat{k}$
$\Rightarrow 3\mu -2\lambda =1\dots (i)$
$4\mu -3\lambda =2\dots (ii)$
On solving Eqs. (i) and (ii), we get
$\mu =-1,\lambda =-2$
Put $\lambda =-2$ in $L_1$
We get $r=\hat{i}+2\hat{j}+3\hat{k}$
$\therefore$ Point of intersection of $L_1$ and $L_2$ is $\hat{i}+2\hat{j}+3\hat{k}$