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).

dk

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:

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

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 \((3,4,5).\)

---

From the distance formula we have:

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

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

---

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 \(a?\)

---

From the distance formula,

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

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

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

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 \(P,Q\) are points on the first and second lines respectively.

Find the distance between the lines \[\frac{x+2}{2}=\frac{y-1}{3}=\frac{z}{1}\] and \[\frac{x-3}{-1}=\frac{y}{1}=\frac{z+1}{2}\]

---

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

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

• Distance formula
• Vectors