Let L be the line passing through the point P4 3 1 with dire

Let L be the line passing through the point P=(4, 3, 1) with direction vector d=[1, 3, 5]^T, and let T be the plane defined by x4y+2z = 18. Find the point Q where L and T intersect.

Solution

parametric equation of line : x = -4 +t ; y = 3 +3t ; z = 1 -5t

Equation of plane : x -4y +2z = -18

Substitute the points (x, y, z):

-4 +t -4(3 +3t)+2(1- 5t) = -18

-4 +t -12 -12t +2 - 10t = -18

-21t -14 = -18

-21t = -4

t = 4/21

So, point of intersection : x = -4 +4/21 ; y = 3 +3(4/21) ; z = 1 -5(4/21)

x = -80/21 ; y = 85/21 ; z = 1/21

Point Q : ( -80/21, 85/21 , 1/21)