Question

Find the coordinates of the point where the line through (3,-4,-5) and (2,-3,1)crosses the plane 2x+y+z=7.

Answer 1

It is known that the equation of the line through the points (x1,y1,z1) and (x2,y2,z2), is
$\frac{x-x_1}{x_2}$-x1=$\frac{y-y_1}{y_2}$-y1=$\frac{z-z_1}{z_2}$-z1

Since the line passes through the points (3,-4,-5) and (2,-3,1), its equation is given by,

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

⇒$\frac{x-3}{-1}$=$\frac{y+4}{1}$=$\frac{z+5}{6}$=k (say)

⇒x=3-k, y=k-4, z=6k-5

Therefore, any point on the line is of the form (3-k, k-4, 6k-5).
This point lies on the plane, 2x+y+z=7

∴2(3-k)+(k-4)+(6k-5)=7
⇒5k-3=7
⇒k=2

Hence, the coordinates of the required point are (3-2, 2-4, 62-5) i.e., (1,-2,7).