## 17Calculus - Vector Lines-Planes Application - Distances Between Points, Lines and Planes

Using Vectors

Applications

### Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

### Articles

On this page we cover a few applications of vectors to calculate distance between points, lines and planes. There are some unique tricks to calculating these distances that you might not think of. Here is a list of topics on this page.

Distance Between a Point and a Line In Space 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.

 Using the dot 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.

### Dr Chris Tisdell - Distance from point to line (using the dot product)

video by Dr Chris Tisdell

 Using the cross product

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.

### MIP4U - Determining the Distance Between a Line and a Point (using the cross product) [1min-17secs]

video by MIP4U

Okay, now let's work some practice problems calculating distances between points and lines.

Practice

Determine the distance between the line $$x=2+3t, y=-4+t, z=-2t$$ and the point $$(1,-6,2)$$.

Problem Statement

Determine the distance between the line $$x=2+3t, y=-4+t, z=-2t$$ and the point $$(1,-6,2)$$.

Solution

### 1848 video

video by MIP4U

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Calculate the distance between the point $$B(1,2,3)$$ and the line $$x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}$$

Problem Statement

Calculate the distance between the point $$B(1,2,3)$$ and the line $$x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}$$

Solution

### 1856 video

video by Dr Chris Tisdell

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Find the distance between the point $$(4,1,-2)$$ and the line $$x=1+t, y=3-2t, z=4-3t$$.

Problem Statement

Find the distance between the point $$(4,1,-2)$$ and the line $$x=1+t, y=3-2t, z=4-3t$$.

Solution

### 1857 video

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Find the distance between the point $$Q(4,4,1)$$ and the line $$x=2t+1, y=-t+7, z=8t+5$$.

Problem Statement

Find the distance between the point $$Q(4,4,1)$$ and the line $$x=2t+1, y=-t+7, z=8t+5$$.

Solution

### 1862 video

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Distance Between Two Parallel Lines

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.

Distance Between a Point and a Plane 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.

Practice

Compute the distance between the point $$B(1,2,3)$$ and the plane $$x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda_1 \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + \lambda_2 \begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix}$$

Problem Statement

Compute the distance between the point $$B(1,2,3)$$ and the plane $$x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda_1 \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + \lambda_2 \begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix}$$

Solution

### 1854 video

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Find the distance between the point $$B(1,2,3)$$ and the plane $$2x - y + z = 5$$

Problem Statement

Find the distance between the point $$B(1,2,3)$$ and the plane $$2x - y + z = 5$$

Solution

### 1855 video

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Find the distance between the point $$(1,-2,4)$$ and the plane $$3x + 2y + 6z = 5$$.

Problem Statement

Find the distance between the point $$(1,-2,4)$$ and the plane $$3x + 2y + 6z = 5$$.

Solution

### 1859 video

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Find the distance between the point $$(7,-3,9)$$ and the plane $$x - 2y + 3z = 6$$.

Problem Statement

Find the distance between the point $$(7,-3,9)$$ and the plane $$x - 2y + 3z = 6$$.

Solution

### 1860 video

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Find the distance between the point $$Q(3,8,-1)$$ and the plane $$4(x+2) - (y+5) + 2z = 0$$.

Problem Statement

Find the distance between the point $$Q(3,8,-1)$$ and the plane $$4(x+2) - (y+5) + 2z = 0$$.

Solution

### 1863 video

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Find the distance between the point $$(0,0,0)$$ and the plane $$2x + y - 2z = 4$$.

Problem Statement

Find the distance between the point $$(0,0,0)$$ and the plane $$2x + y - 2z = 4$$.

Solution

### 1864 video

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Distance Between Two Parallel Planes

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.

Practice

Find the distance between the parallel planes $$2x - 3y + z = 4$$ and $$4x - 6y + 2z = 3$$.

Problem Statement

Find the distance between the parallel planes $$2x - 3y + z = 4$$ and $$4x - 6y + 2z = 3$$.

Solution

### 1858 video

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Find the distance between the parallel planes $$-5x + 3y - 2z = -14$$ and $$-5x + 3y - 2z = -25$$.

Problem Statement

Find the distance between the parallel planes $$-5x + 3y - 2z = -14$$ and $$-5x + 3y - 2z = -25$$.

Solution

### 1861 video

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