Question

The point of intersection of the lines
$l_1=\overrightarrow{r}\left(t\right)=(\overrightarrow{i}-6\overrightarrow{j}+2\overrightarrow{k})+t(\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})$
$l_2=\overrightarrow{R}\left(u\right)=(4\overrightarrow{j}+\overrightarrow{k})+u(2\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k})$ is

• A. $(4,4,5)$
• B. $(6,4,7)$
• C. $(8,8,9)$
• D. $(10,12,11)$

Answer 1

The correct option is C $(8,8,9)$

Any point on $l_1$ is $(1+t,-6+2t,2+t)$

Any point on $l_2$ is $(2u,4+u,1+2u)$

Point of intersection will lie on both so,

$1+t=2u$

$\Rightarrow t=2u-1$ --------(1)

and

$-6+2t=4+u$

$\Rightarrow -6+2(2u-1)=4+u$

$\Rightarrow -6+4u-2=4+u$

$\Rightarrow 3u=12$

$\Rightarrow u=4$

$\therefore t=7$

$\therefore$ Point of intersection is $(1+7,-6+14,2+7)$

i.e., $(8,8,9)$