Question

Consider the plane P with equation −2x−2y−z = −11 and line L with equations
x = −3−4t
y = −2−3t
z = −6−4t
Find two distinct points Q1 and Q2 on the line L at distance 3 to the plane P.

Q1 = (0, 0, 0)


Q2 = (0, 0, 0)

Answer 1

Let Q be a general point on the line
Q is (-3-4t, -2-3t, -6-4t)
Plane : -2x -2y -z + 11 = 0

Using the distance from a point to a plane formula
| -2(-3-4t) - 2(-2-3t) - (-6-4t) + 11|/ √(4+4+1) = 3
± (6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
using the positive
18t + 27 = 9
18t = -18
t = -1
then Q is (-3+4, -2+3, -6+4) ,

<b>Q1 is (1, 1, -2)</b>

using the negative result from above:
-(6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
-18t - 27 = 9
-18t = 36
t = -2

Q2 is (-3+8, -2+6, -6+8)

= <b>Q2(5, 4, 2)</b>


check:
midpoint of Q1Q2 = (3, 5/2, 0) , this point should lie on our plane
LS = -2(3) -2(5/2) - (0)
= -6 -5 - 0 = -11
RS = -11

looks good!