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Intersection of a line and a plane

A line in space is written with a parameter: a point plus the parameter times the direction. To find the intersection with a plane you insert the three expressions into the equation of the plane. Then only the parameter remains, and you solve for it. You get the point by putting the parameter value back into the line. If the coefficient of the parameter is zero, the line is parallel to the plane.

x = x_0 + at,\ y = y_0 + bt,\ z = z_0 + ct

the line

Symbols

$t$	the parameter
$a,\ b,\ c$	the direction

Example

The line $[1 + t, 2 - t, 3t]$ and the plane $x + y + z = 6$ : $(1 + t) + (2 - t) + 3t = 3 + 3t = 6$ , so $t = 1$ . The point has $x = 2$ , $y = 1$ and $z = 3$ .

Calculate the parameter first. The coordinates of the point come afterwards, not the other way around.

Practise lines and planes in space for free →

← Distance from a point to a plane · Volume of a parallelepiped →

Part of Mathematics R2: Lines and planes in space.