Using the figure on the right, we will show how to find the distance d between the line L and the point Q. There are several ways to do this. We will look at two in particular, one using the dot product, the other using the cross product.

1. Find a vector along the line L. This vector does not necessarily have to be a unit vector.
2. Find a vector from any point P on the line to the point Q.
3. Calculate the projection of the vector \( \overrightarrow{PQ}\) onto \(\vec{u}\), \(proj_{\vec{u}}\overrightarrow{PQ}\).
4. Find \( \overrightarrow{PQ} - proj_{\vec{u}}\overrightarrow{PQ} \)
5. Calculate \( d = \| \overrightarrow{PQ} - proj_{\vec{u}}\overrightarrow{PQ} \| \)

Here is a great video explaining this in more detail and showing an example.

The idea is to determine a point on the line and call it P (Any point will do) which we use to find the vector \(\overrightarrow{PQ}\). Then we find a unit vector \(\vec{u}\) along the line.

Now we can use \(\vec{u}\) and \( \overrightarrow{PQ}\) to determine the sine of the angle \(\theta\) as \(\displaystyle{ \sin(\theta) = \frac{d}{\| \overrightarrow{PQ} \|} ~~~ \to ~~~ d = \| \overrightarrow{PQ} \| \sin(\theta) }\)

From a geometric property of the cross product we know that \(\displaystyle{ \| \overrightarrow{PQ} \times \vec{u} \| = \| \overrightarrow{PQ} \| ~ \| \vec{u} \| \sin(\theta) }\) where \( \theta \) is the angle between the vectors. So \(\displaystyle{ \sin(\theta) = \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{\| \overrightarrow{PQ} \| ~ \| \vec{u} \|} }\). Substituting \(\sin(\theta) \) from this equation into the equation for d we get

\(\displaystyle{ d = \| \overrightarrow{PQ} \| \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{\| \overrightarrow{PQ} \| ~ \| \vec{u} \|} = \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{ \| \vec{u} \|}. }\)

And since \(\vec{u}\) is a unit vector, it's length is one, so we can write the equation for \(d \) as \(\displaystyle{ d = \| \overrightarrow{PQ} \times \vec{u} \|. }\)

Here is a video clip explaining this derivation.

To find the distance between two parallel lines, we extend the logic of the previous section by first finding one point on one of the lines (any point will do) and then finding the distance between that point and the other line.

The goal is to find the distance between a point in space and a plane. In the figure on the right, we want to find the distance d, the perpendicular distance between Q and the plane. As an overview, here is how we will do this. We will find a unit normal vector to the plane at some point P. We then project the vector \( \overrightarrow{PQ} \) onto the unit normal vector, labeled \(proj\) in the figure on the right. This distance is equal to the distance d. Spend a few minutes going over this discussion and looking at the figure to make sure you understand this.

We use the dot product to get the projection of one vector onto another. The equation is \(\displaystyle{ d = \left| \overrightarrow{PQ} \cdot \frac{\vec{n}}{\| \vec{n} \|} \right| = \frac{\left| \overrightarrow{PQ} \cdot \vec{n} \right|}{\| \vec{n} \|} }\)

Okay, time for some practice problems calculating the distance between a point and a plane.

To find the distance between two parallel planes, we extend the logic of the previous section by first finding one point on one of the planes (any point will do) and then finding the distance between that point and the other plane.