Question

Find the coordinates of the point where the line through the points $(3,-4,-5)$ and $(2,-3,1)$, crosses the plane determined by thee points $(1,2,3),(4,2,-3)$ and $(0,4,3)$.

Answer 1

Given that the plane passes through the points $A(1,2,3)$ , $B(4,2,-3)$ and $C(0,4,3)$

$BA=3\hat{i}-6\hat{k}$ , $CA=-\hat{i}+2\hat{j}$

The Vector $n=BA\times CA=12i+6j+6k$ will be perpendicular to the plane passing through points $A,B,C$

So the directional ratios of plane will be $12,6,6$

The equation of plane will be $12x+6y+6z=d$ and it passes through point $a=A(1,2,3)$

So the equation of plane will be $\overrightarrow{r}\cdot \hat{n}=\overrightarrow{a}.\hat{n}$

i.e.$2x+y+z=7$

The equation of line passing through points $D(3,-4,-5)$ and $E(2,-3,1)$

is $\frac{x-3}{1}=\frac{y+4}{-1}=\frac{z+5}{-6}$

So the point on the line will be in the form of $(t+3,-4-t,-5-6t)$ and this point will lie on the plane.

So we have $2t+6-4-t-5-6t=7$

$\Rightarrow t=-2$

So the point where the line crosses the plane will be $(1,-2,7)$