Perpendicular lines in 3 space?

Question

A line passes through A = (1, 4, 1) and is perpendicular to the plane 5x+5y+2z=2.
Find the point P of intersection of the line and the plane.
Thanks for the help!

Answer 1

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

Find the point P of intersection of the line and the plane.

The normal vector n, of the plane
5x + 5y + 2z - 2 = 0

can be taken from the coefficients.

n = <5, 5, 2>

This vector is also the directional vector of the line thru
A(1, 4, 1) that is perpendicular to the plane.

The equation of the line is:

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

The value of the line and the plane will be equal at the point of intersection. Plug in the line values for x, y, and z into the plane and solve for t.

5x + 5y + 2z - 2 = 0
5(1 + 5t) + 5(4 + 5t) + 2(1 + 2t) - 2 = 0
5 + 25t + 20 + 25t + 2 + 4t - 2 = 0
25 + 54t = 0
54t = -25
t = -25/54

x = 1 + 5t = 1 - 5(25/54) = -71/54
y = 4 + 5t = 4 - 5(25/54) = 91/54
z = 1 + 2t = 1 - 2(25/54) = 4/54 = 2/27

The point of intersection is P(-71/54, 91/54, 2/27).