### Finding where a line intersects a plane

**Problem**

Where does the line **r** = (1, 2, -5) + t (2, -3, 1) meet the
plane 2x + 5y - 3z = 6?
**Solution**

Letting **r** = (x, y, z) we have

(x, y, z) = (1, 2, -5) + t (2, -3, 1).
For points on this line, the components of their position vectors are
given by

x = 1 + 2t

y = 2 - 3t

z = -5 + t
Also, we know that 2x + 5y - 3z = 6 for points lying in the plane.

For a point in common, all these equations must be simultaneously
satisfied. Substituting for x, y and z in the 4th equation gives

2 (1 + 2t) + 5 (2 - 3t) - 3 (-5 + t) = 6.
This gives -14t = -21 so t = 3/2. Substituting this value
of t gives x = 4, y = -5/2 and z = -7/2.

The position vector of the point where the line meets the plane
is ( 4, -5/2, -7/2).

Check for yourself that this point does lie on the plane 2x + 5y - 3z = 6.
(Doing this gives a quick check against arithmetical slips!)
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