>Intersections>How to Find the Intersection Between Line and Plane

How to Find the Intersection Between Line and Plane

The intersection between a line and a plane is either one point in the plane or an infinite number of points. In the case of an infinite number of points, the line actually lies in the plane.

The intersection between a line and a plane

To find the intersection between a line and a plane, you can follow this recipe:

Rule

Intersection Between a Line and a Plane

1.: Put the parametric equation of the line into the equation of the plane.
2.: You are now left with an expression with only one variable, $t$ .
3.: Solve for $t$ .
4.: Put this $t$ -value back into the parametric equation for the line and find the intersection $(x, y, z)$ .

Example 1

You want to find the intersection between the line
$\begin{array}{l} l : x (t) & = 1 + t, \\ y (t) & = 2 t, \\ z (t) & = 2 + t \end{array}$

$l : x (t) = 1 + t, y (t) = 2 t, z (t) = 2 + t$

and the plane
$α : x - 3 y + 2 z = 9 .$

Items 1 to 3.

In this problem, it’s natural to execute all three of these steps at the same time. Look through the example to convince yourself that this is the case! You insert the parametric equation of the line into the equation of the plane, and solve for $t$ . $\begin{array}{l} 1 + t - 3 (2 t) + 2 (2 + t) & = 9 \\ 1 + t - 6 t + 4 + 2 t & = 9 \\ - 3 t & = 4 \\ t & = \frac{- 4}{3} \end{array}$

Item 4.

Now you can put this back into the parametric equation for the line:

\begin{array}{l} x (\frac{- 4}{3}) & = 1 + \frac{- 4}{3} = \frac{- 1}{3}, \\ y (\frac{- 4}{3}) & = 2 (\frac{- 4}{3}) = \frac{- 8}{3}, \\ z (\frac{- 4}{3}) & = 2 + \frac{- 4}{3} = \frac{2}{3} \end{array}

That means the intersection is at the point

P = (\frac{- 1}{3}, \frac{- 8}{3}, \frac{2}{3}) .

Note! If you cannot find a value for $t$ using this method, there are two possible reasons: Either the line lies in the plane, or it is parallel to the plane.

If the line lies in the plane, you will end up with a solution like $0$ = $0$ or something similar. This is because all $t$ -values will give you a point in the plane.

In any case, you will have to check whether this is true to convince yourself that you haven’t made any mistakes. You can do so by checking whether the directional vector of the line is perpendicular to the normal vector of the plane. If they aren’t perpendicular, there has to be one intersecting point.

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