Find the intersection of a line x= -1+t, y= 3+2t, z= -t, with the plane -4x + y - 2z - 7 =0?

Question

In a more cleared form:

Find the intersection of the line x= -1+t, y=3+2t, z=-t with the plane -4x + y - 2z - 7= 0.

Answer 1

(Substitute the x, y and z equation into the equation of the plane, solve for t, then plug then back to the x, y and z equation.)

Plane equation: -4x + y - 2z - 7= 0
Line equations: x= -1+t, y=3+2t, z=-t

-4(-1+t) + (3+2t) - 2(-t) - 7 = 0

4 - 4t + 3 + 2t +2t - 7 = 0

0 = 0

=> Infinitely many solutions

=> the line is on the given plane


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As a side note, there are 3 possibilities for this kind of question.

1) Solution: 0 = 0

Infinitely many solution, which means line is on the plane

2) Solution: t = C (C = any real number)

One unique solution, which means line intersect the plane on exactly 1 point.

3) Solution C = 0 (C = any non zero real number)

No solution (Since any non zero real number is not equal to 0), which means the line will never intersect the plane. In other words, the line is parallel to the plane.

Answer 2

Find the intersection of the line x = -1+t, y = 3+2t, z = -t with the plane -4x + y - 2z - 7 = 0.

The line and plane are equal at the point of intersection. Plug in the value of (x, y, z) at the point of intersection in terms of t.

-4x + y - 2z - 7 = 0
-4(-1 + t) + (3 + 2t) - 2(-t) - 7 = 0
4 - 4t + 3 + 2t + 2t - 7 = 0
0 = 0

This is true for all t. Therefore the line lies in the plane. The intersection of the line and plane is the line.