A line passes through A = (2, 1, 3) and is perpendicular to the plane 3x+4y+5z=7. Find Intersection.?

Question

What is the point of intersection of the Line and Plane?

Answer 1

The normal vector n, to the plane 3x + 4y + 5z - 7 = 0 is:
n = <3, 4, 5>

The directional vector to the line thru A(2, 1, 3) that is perpendicular to the plane is also n.

The equation of the line L is:

L = A + tn
L = <2, 1, 3> + t<3, 4, 5>
L = <2 + 3t, 1 + 4t, 3 + 5t>
where t is a constant that ranges over the real numbers

The line and plane intersect in a common point. Set them equal and solve for t.

3x + 4y + 5z - 7 = 0
3(2 + 3t) + 4(1 + 4t) + 5(3 + 5t) - 7 = 0
6 + 9t + 4 + 16t + 15 + 25t - 7 = 0
18 + 50t = 0
50t = -18
t = -18/50 = -9/25

x = 2 + 3t = 2 - 27/25 = 23/25
y = 1 + 4t = 1 - 36/25 = -11/25
z = 3 + 5t = 3 - 45/25 = 30/25 = 6/5

The point of intersection is (23/25, -11/25, 6/5).