How to Find the Distance Between a Line and a Plane

A line l parallel to a plane

If a line and a plane don’t intersect, they will always be parallel. Meaning, the distance between the line and the plane always stays the same. To check if the line and the plane are parallel to each other, you can take the dot product of the directional vector of the line and the normal vector to the plane. If the dot product is zero, the line does not intersect with the plane. To find the distance between a line and a plane, you just pick a point on the line and use the equation for the distance between a point and a plane. You can use this method:

Rule

Distance Between Line and Plane

1.

Let

P = (x_{1}, y_{1}, z_{1})

be a point on the line

l

and let

a x + b y + c z + d = 0

be the equation of the plane $α$ . Then ${\vec{n}}_{α} = (a, b, c)$ is a normal vector to the plane $α$ .

2.

Put the values into the formula for the distance from a point to a plane to find the distance.

Example 1

You have the line
$\begin{array}{l} l : x (t) & = 1 + t, \\ y (t) & = 3 t, \\ z (t) & = 1 + 4 t \end{array}$

$l : x (t) = 1 + t, y (t) = 3 t, z (t) = 1 + 4 t$

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

Find the distance between them.

1.: If you set $t = 0$ , you get that $P = (1, 0, 1)$ is a point on $l$ . A normal vector to $α$ is $\vec{n} = (1, - 3, 2)$ .
2.: When you insert this into the formula, you get $\begin{array}{l} D & = \frac{| 1 \cdot 1 - 3 \cdot 0 + 2 \cdot 1 - 9 |}{\sqrt{1^{2} + {(- 3)}^{2} + 2^{2}}} \\ = \frac{| 1 + 0 + 2 - 9 |}{\sqrt{1 + 9 + 4}} \\ = \frac{| - 6 |}{\sqrt{14}} \\ = \frac{6}{\sqrt{14}} . \end{array}$

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