Question

Find the point at which $\overrightarrow{r}=(\overrightarrow{i}+2\overrightarrow{j}-5\overrightarrow{k})+t(2\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k})$ meets the plane $\overrightarrow{r}\cdot (2\overrightarrow{i}+4\overrightarrow{j}-\overrightarrow{k})=3$

Answer 1

The equation of the straight line in the cartesian form is
$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=d$ (say)
$\therefore \frac{x-1}{2}=d\Rightarrow x-1=2d\Rightarrow x=2d+1$
$\frac{y-2}{-3}=d\Rightarrow y-2=-3d\Rightarrow y=2-3d=-3d+2$
$\frac{z+5}{4}=d\Rightarrow z+5=4d\Rightarrow z=4d-5$
$\therefore$ Any point on this line is of the form
$(2d+1,-3d+2,4d-5)$ .... (1)
The cartesian equation of the given plane is $2x+4y-z-3=0$.
The point $(2d+1,-3d+2,4d-5)$ lies in this plane.
$\therefore 2(2d+1)+4(-3d+2)-(4d-5)-3=0$
$\Rightarrow 4d+2-12d+8-4d+5-3=0$
$\Rightarrow -12d+12=0$
$\Rightarrow -12d=-12$

$\Rightarrow d=1$
Substituting $d=1$ in equation (1) in we get, the required point as $(3,-1,-1)$.