How to Find the Distance from a Point to a Line in 3D?

Finding the distance from a point to a line in 3D space can be a bit tricky, but with the right approach, it becomes manageable. We use vector projection and the cross product to solve this problem.

Step-by-Step Solution

Given Information

Suppose you have a point $P(x_1, y_1, z_1)$ and a line defined by a point $A(x_2, y_2, z_2)$ and a direction vector $mathbf{d} = langle a, b, c rangle$

1. Vector from Point to Line

First, find the vector from point $A$ on the line to point $P$:

$mathbf{AP} = langle x_1 – x_2, y_1 – y_2, z_1 – z_2 rangle$

2. Cross Product

Next, compute the cross product of $mathbf{AP}$ and the direction vector $mathbf{d}$:

$mathbf{AP} times mathbf{d} = left| begin{matrix} mathbf{i} & mathbf{j} & mathbf{k} \ x_1 – x_2 & y_1 – y_2 & z_1 – z_2 \ a & b & c end{matrix} right|$

This results in a new vector $mathbf{N}$:

$mathbf{N} = langle (y_1 – y_2)c – (z_1 – z_2)b, (z_1 – z_2)a – (x_1 – x_2)c, (x_1 – x_2)b – (y_1 – y_2)a rangle$

3. Magnitude of the Cross Product

Find the magnitude of vector $mathbf{N}$:

$|mathbf{N}| = sqrt{[(y_1 – y_2)c – (z_1 – z_2)b]^2 + [(z_1 – z_2)a – (x_1 – x_2)c]^2 + [(x_1 – x_2)b – (y_1 – y_2)a]^2}$

4. Magnitude of the Direction Vector

Find the magnitude of the direction vector $mathbf{d}$:

$|mathbf{d}| = sqrt{a^2 + b^2 + c^2}$

5. Distance Formula

Finally, the distance $D$ from the point $P$ to the line is given by:

$D = frac{|mathbf{N}|}{|mathbf{d}|}$

Example Calculation

Let’s say point $P$ is $(1, 2, 3)$ and the line passes through point $A(4, 5, 6)$ with direction vector $mathbf{d} = langle 1, 0, 1 rangle$

1. Calculate $mathbf{AP}$:

$mathbf{AP} = langle 1 – 4, 2 – 5, 3 – 6 rangle = langle -3, -3, -3 rangle$

2. Compute $mathbf{AP} times mathbf{d}$:

$mathbf{AP} times mathbf{d} = left| begin{matrix} mathbf{i} & mathbf{j} & mathbf{k} \ -3 & -3 & -3 \ 1 & 0 & 1 end{matrix} right| = langle -3, 0, 3 rangle$

3. Find $|mathbf{N}|$:

$|mathbf{N}| = sqrt{(-3)^2 + 0^2 + 3^2} = sqrt{18} = 3sqrt{2}$

4. Find $|mathbf{d}|$:

$|mathbf{d}| = sqrt{1^2 + 0^2 + 1^2} = sqrt{2}$

5. Calculate the distance $D$:

$D = frac{3sqrt{2}}{sqrt{2}} = 3$

So, the distance from point $P(1, 2, 3)$ to the line is 3 units.

Conclusion

By following these steps, you can determine the distance from any point to a line in 3D space. Understanding this process can be very useful in fields like physics, engineering, and computer graphics.

3. Brilliant – Distance from Point to Line

Citations

1. 1. Khan Academy – Distance between point and line
2. 2. Wolfram MathWorld – Point-Line Distance