Intersection of a Line and a Plane

Here you will learn how to find intersection of a line and a plane with examples.

Let’s begin – 

Intersection of a Line and a Plane

Let the equation of a line be $x – x_1\over l$ = $y – y_1\over m$ = $z – z_1\over n$ and that of a plane be ax + by + cz + d = 0.

The coordinates of any point on the line  $x – x_1\over l$ = $y – y_1\over m$ = $z – z_1\over n$ is given 

$x – x_1\over l$ = $y – y_1\over m$ = $z – z_1\over n$ = r (say)

or, $(x_1 + lr, y_1 + mr, z + nr)$                …………(i)

If it lies on the plane ax + by + cz + d = 0, then

$a(x_1 + lr)$ + $b(y_1 + mr)$ + $c(z_1 + nr)$ + d = 0

$\implies$ $(ax_1 + by_1 + cz_1 + d)$ + r(al + bm + cn) = 0

$\implies$ r = -$(ax_1 + by_1 + cz_1 + d)\over al + bm + cn$

Substituting the value of r in (i), we obtain the coordinates of the required point of intersection.

In order to find the coordinates of the point of intersection of a line and a plane, we may use the following algorithm,

Algorithm :

1). Write the coordinates of any point on the line in terms of some parameters r (say).

2). Substitute these coordinates in the equation of the plane to obtain the value of r.

3). Put the value of r in the coordinates of the point in step 1.

Example : Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.

Solution : The equation of the line passing through A and B is

$x – 3\over 5 – 3$ = $y – 4\over 1 – 4$ = $z – 1\over 6 – 1$

or,  $x – 3\over 2$ = $y – 4\over -3$ = $z – 1\over 5$

The coordinates of any point on this line are given by

$x – 3\over 2$ = $y – 4\over -3$ = $z – 1\over 5$ = $\lambda$ 

$\implies$ x = $2\lambda + 3$,  y = $-3\lambda + 4$,  z = $5\lambda + 1$

So, $(2\lambda + 3, -3\lambda + 4, 5\lambda + 1)$ are coordinates of any point on the line passing through A and B. If it lies on XY-plane i.e z = 0.Then.

$5\lambda + 1$ = 0 $\implies$ $\lambda$ = -$1\over 5$

Thus, the coordinates of the required point are (13/5, 23/5, 0).