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 the points $(1,2,3),(4,2,-3)$ and $(0,4,3)$.

Answer 1

The line through $(3,-4,-5)$ and $(2,-3,1)$ is given by

$\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$

Or $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$ ...$\left(i\right)$

The plane determined by the points $(1,2,3),(4,2,-3)$ and $(0,4,3)$ is:

$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 4-1& 2-2& -3-3\\ 0-1& 4-2& 3-3\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y-2& z-3\\ 3& 0& -6\\ -1& 2& 0\end{array}\right|=0$

$\Rightarrow (x-1)(0+12)-(y-2)(0-6)+(z-3)(6-0)=0$

$\Rightarrow 12(x-1)+6(y-2)+6(z-3)=0$

$\Rightarrow$ $12x+6y+6z-42=0$

or $2x+y+z-7=0$ ...$\left(ii\right)$

$P(-\mu +3,\mu -4,6\mu -5)$ is the general point for line $\left(i\right)$

If this point lies on plane $\left(ii\right)$, we have

$-2\mu +6+\mu -4+6\mu -5-7=0\Rightarrow \mu =2$

$\therefore$ $P(1,-2,7)$ is the point of intersection.