Question

Find the ratio in which the line joining the points (2, 4, 16) and (3, 5, -4) is divided by the plane 2x – 3y+ z+ 6 = 0. Also find the co-ordinates of the point of division

Answer 1

Answer:

Step-by-step explanation:

let the plane intersects the join of points in the ratio k:1

let (x,y,z) be the point of intersection.

$x=\frac{3k+2}{k+1} \\y=\frac{5k+4}{k+1} \\z=\frac{-4k+16}{k+1} \\\because ~(x,y,z)~lies~on~the~plane.\\2(\frac{3k+2}{k+1} )-3(\frac{5k+4}{k+1} )+\frac{-4k+16}{k+1} +6=0\\multiply~by~k+1\\2(3k+2)-3(5k+4)+(-4k+16)+6(k+1)=0\\6k+4-15k-12-4k+16+6k+6=0\\-7k+14=0\\k=2\\x=\frac{3*2+2}{2+1} =\frac{8}{3} \\y=\frac{5*2+4}{2+1}= \frac{14}{3} \\z=\frac{-4*2+16}{2+1} =\frac{8}{3}$

point of intersection is (8/3,14/3,8/3)

and ratio of division is 2:1