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

Equation of plane determined by the points (1,2,3),(4,2,-3) and (0,4,3):

$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 4-1& 2-2& -3-3\\ 0-1& 4-2& 3-3\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x-1& y-2& z-3\\ 3& 0& -6\\ -1& 2& 0\end{array}\right|=0$

On expanding along $R_1$,we get: 2x+y+z=7...(i)

Now erquation of line through the points (3,-4,-5) and (2,-3,1) is :$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$

Coordinates of random point on line:$P(-\lambda +3,\lambda -4,6\lambda -5)$

For the point of intersection of line and plane,point P must satisfy (i),so we get:$\lambda =2$

Hence the required point of intersection is :P(1,-2,7).