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.
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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.
video by Dr Chris Tisdell |
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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.
video by MIP4U |
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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 |
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Determine the distance between the line \( x=2+3t, y=-4+t, z=-2t \) and the point \( (1,-6,2) \).
Solution |
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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 |
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Calculate the distance between the point \( B(1,2,3) \) and the line \(x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}\)
Solution |
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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 |
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Find the distance between the point \( (4,1,-2) \) and the line \( x=1+t, y=3-2t, z=4-3t \).
Solution |
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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 |
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Find the distance between the point \( Q(4,4,1) \) and the line \( x=2t+1, y=-t+7, z=8t+5 \).
Solution |
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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 |
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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 |
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Find the distance between the point \( B(1,2,3) \) and the plane \( 2x - y + z = 5 \)
Problem Statement |
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Find the distance between the point \( B(1,2,3) \) and the plane \( 2x - y + z = 5 \)
Solution |
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Find the distance between the point \( (1,-2,4) \) and the plane \( 3x + 2y + 6z = 5 \).
Problem Statement |
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Find the distance between the point \( (1,-2,4) \) and the plane \( 3x + 2y + 6z = 5 \).
Solution |
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Find the distance between the point \( (7,-3,9) \) and the plane \( x - 2y + 3z = 6 \).
Problem Statement |
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Find the distance between the point \( (7,-3,9) \) and the plane \( x - 2y + 3z = 6 \).
Solution |
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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 |
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Find the distance between the point \( Q(3,8,-1) \) and the plane \( 4(x+2) - (y+5) + 2z = 0 \).
Solution |
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Find the distance between the point \( (0,0,0) \) and the plane \( 2x + y - 2z = 4 \).
Problem Statement |
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Find the distance between the point \( (0,0,0) \) and the plane \( 2x + y - 2z = 4 \).
Solution |
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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 |
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Find the distance between the parallel planes \( 2x - 3y + z = 4 \) and \( 4x - 6y + 2z = 3 \).
Solution |
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Find the distance between the parallel planes \( -5x + 3y - 2z = -14 \) and \( -5x + 3y - 2z = -25 \).
Problem Statement |
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Find the distance between the parallel planes \( -5x + 3y - 2z = -14 \) and \( -5x + 3y - 2z = -25 \).
Solution |
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Really UNDERSTAND Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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