3D Coordinate Geometry - Distance

In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition.

In three-dimensional space, points are represented by their positions along the \(x\)-, \(y\)-, and \(z\)-axes, which are each perpendicular to one another; this is analogous to the 2d coordinate geometry interpretation in which each point is represented by only two coordinates (along the \(x\)- and \(y\)-axes).

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In the figure above, the goal is to find the distance from the point \(\left(x_{1},y_{1},z_{1}\right)\) to the point \(\left(x_{2},y_{2},z_{2}\right).\) From the distance formula in two dimensions, the length of the the yellow line is

\[\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }{ -y }_{ 1 }) }^{ 2 } }. \]

Let \(d\) be the distance from the point \(\left(x_{1},y_{1},z_{1}\right)\) to \(\left(x_{2},y_{2},z_{2}\right)\) (the red line, and the desired distance). Then, using the Pythagorean theorem,

\[\begin{align} {d }^{ 2 }&= \left(\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+({ { y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \right)^{ 2 }+{ ({ z }_{ 2 }-{ z }_{ 1 }) }^{ 2 }\\ \Rightarrow d&=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+({ { y }_{ 2 }-{ y }_{ 1 }) }^{ 2 }+{ ({ z }_{ 2 }-{ z }_{ 1 }) }^{ 2 } }. \end{align}\]

The above equation is the general form of the distance formula in 3D space. A special case is when the initial point is at the origin, which reduces the distance formula to the form

\[d=\sqrt { { { x } }^{ 2 }+{ { y } }^{ 2 }+{ { z } }^{ 2 } },\]

where \((x,y,z)\) is the terminal point. This equation extends the distance formula to 3D space.

Find the distance between the points \((2,-5,7)\) and \((3,4,5).\)

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From the distance formula, we have

\[\begin{align} d&=\sqrt { { (3-2) }^{ 2 }+\big({ 4-(-5)\big) }^{ 2 }+{ (5-7) }^{ 2 } } \\ &=\sqrt { 1+81+4 } \\ &=\sqrt { 86 }.\ _\square \end{align}\]

How far is the point \(P=(3,4,5)\) from the origin?

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Since the second point is the origin, or \((0,0,0)\), the distance is

\[d=\sqrt{3^2+4^2+5^2}=5\sqrt{2}.\ _\square\]

If the distance between the two points \((2,0,3)\) and \((-4,a,-1)\) is 8, what is the value of \(a?\)

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From the distance formula,

\[\begin{align*} d&=\sqrt{\big(2-(-4)\big)^2+(0-a)^2+\big(3-(-1)\big)^2}\\ &=\sqrt{36+a^2+16}\\ &=\sqrt{52+a^2}\\ &=8. \end{align*} \]

Therefore,

\[\begin{align*} 52+a^2&=8^2\\&=64\\ a&=\pm2\sqrt{3}.\ _\square \end{align*}\]

Determining the distance between a point and a plane follows a similar strategy to determining the distance between a point and a line. Consider a plane defined by the equation

\[ax + by + cz + d = 0\]

and a point \((x_0, y_0, z_0)\) in space. Then the normal vector to the plane is

\[\mathbf{v} = \begin{pmatrix}a\\b\\c\end{pmatrix}\]

and the vector from an arbitrary point on the plane \((x,y,z)\) to the point is

\[\mathbf{w} = \begin{pmatrix}x_0-x\\y_0-y\\z_0-z\end{pmatrix}.\]

The distance from the point to the plane is the projection from \(\mathbf{w}\) onto \(\mathbf{v}\), or

\[ \begin{align*} D &= |\text{proj}_{\mathbf{v}}\mathbf{w}| \\ &= \frac{|\mathbf{v} \cdot \mathbf{w}|}{|\mathbf{v}|} \\ &= \frac{|a(x_0-x)+b(y_0-y)+c(z_0-z)|}{\sqrt{a^2+b^2+c^2}} \\ &= \frac{|ax_0+by_0+cz_0-(ax+by+cz)|}{\sqrt{a^2+b^2+c^2}} \\ &= \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}. \end{align*} \]

The strategy behind determining the distance between 2 skew lines is to find two parallel planes passing through each line; this is because the distance between two planes is easy to calculate using vector projection. Furthermore, the normal vector to these 2 planes can be calculated using the cross product of the vectors representing the direction of the two lines.

In particular, suppose the two lines travel in the directions

\[\begin{align} \mathbf{v}&=\begin{pmatrix}x_1&y_1&z_1\end{pmatrix}\\ \mathbf{w}&=\begin{pmatrix}x_2&y_2&z_2\end{pmatrix}, \end{align}\]

then the normal to the planes can be calculated as

\[\mathbf{n} = \mathbf{v} \times \mathbf{w} = \text{det}\begin{pmatrix}\hat{i}&\hat{j}&\hat{k}\\x_1&y_1&z_1\\x_2&y_2&z_2\end{pmatrix}\]

and the distance between the two planes--which is the same as the distance between the two lines--can be calculated by projecting \(\mathbf{n}\) onto \(PQ\), where \(P\) and \(Q\) are points on the first and second lines, respectively.

Find the distance between the lines

\[\frac{x+2}{2}=\frac{y-1}{3}=\frac{z}{1}\quad \text{and}\quad \frac{x-3}{-1}=\frac{y}{1}=\frac{z+1}{2}.\]

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Following the above strategy, the first line travels in the direction

\[\mathbf{v}=\begin{pmatrix}2&3&1\end{pmatrix}\]

and the second in the direction

\[\mathbf{w}=\begin{pmatrix}-1&1&2\end{pmatrix}.\]

Then

\[\mathbf{n} = \mathbf{v} \times \mathbf{w} = \text{det}\begin{pmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&1\\-1&1&2\end{pmatrix}=\begin{pmatrix}5&-5&5\end{pmatrix}.\]

Since \((-2, 1, 0)\) and \((3, 0, -1)\) are points on the two lines, respectively, the vector \(PQ\) is \(\begin{pmatrix}5&-1&-1\end{pmatrix}\). Then

\[D=\big|\text{proj}_{\mathbf{n}}PQ\big|=\frac{\big|\mathbf{n} \cdot PQ\big|}{|\mathbf{n}|}=\frac{25}{5\sqrt{3}}=\frac{5\sqrt{3}}{3}.\ _\square\]

• Distance Formula
• Vectors