The Midpoint and Distance Formulas in 3D
The Midpoint and Distance Formulas in 3D

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3D Distance Formula

Before learning 3d distance formula, let us recall the 2d distance formula, which gives the length of the line segment that joins two points on a plane. In the same way, the 3d distance formula gives the length of the line segment joining two points in the three-dimensional space. The 2d distance formula for any two points (x1, y1) and (x2, y2) is given as:

\(d = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2}\)

The 3d distance formula is just an extension of this formula. Let us learn this formula along with a few solved examples.

What Is 3D Distance Formula?

The 3d distance formula says, the distance between two points in the space A(x1, y1, z1) and B(x2, y2, z2) is given as:

\(d = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2+(z_2-z_1)^2}\)

Here, d is the distance between A and B or the length of the line segment joining A and B.

Example

Find the distance between the points (1, 3, -2) and (2, 4, -1).

Solution

It is given that

(x1, y1, z1) = (1, 3, -2)

(x2, y2, z2) = (2, 4, -1)

Substitute these values in the 3d distance formula:

\[ \begin{align} d &= \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2+(z_2-z_1)^2}\\[0.2cm] &= \sqrt{(2-1)^2+(4-3)^2+(-1-(-2))^2}\\[0.2cm] &= \sqrt{1+1+1}\\[0.2cm] &= \sqrt{3} \end{align} \]

Thus, the distance between the given two points is \(\sqrt{3}\) units.

Note: The distance between two points, as it is the length of the line segment, is never negative.

Solved Examples Using 3D Distance Formula

  1. Example 1

    Use the 3D distance formula to prove the following points are collinear.

    A(-1, 0, 7), B(3, 2, 1), and C(5, 3, -2).

    Solution:

    Using the 3d distance formula,

    \[ \begin{align} AB &= \sqrt{(3+1)^2+(2-0)^2+(1-7)^2}=2 \sqrt{14}\\[0.2cm] BC &= \sqrt{(5-3)^2+(3-2)^2+(-2-1)^2}= \sqrt{14}\\[0.2cm] CA &= \sqrt{(5+1)^2+(3-0)^2+(-2-7)^2} = 3 \sqrt{14} \end{align}\]

    Here, AB + BC = CA.

    Hence, A, B, and C are collinear.

    Answer: Since AB + BC = CA, the given points are collinear.

  2. Example 2:

    Find a point on the y-axis that is equidistant from the points (−1, 2, 0) and (2, 3, -1)

    Solution:

    We know that the x and z coordinates of any point on the y-axis are 0.

    Hence, we assume the point that is equidistant from the given points to be (0, k, 0). i.e.,

    Distance between (0, k, 0) and (−1, 2, 0) = Distance between (0, k, 0) and (2, 3, -1)

    \(\begin{aligned} \sqrt{(-1-0)^2+(2-k)^2+(0-0)^2}&= \sqrt{(2-0)^2+(3-k)^2+(-1-0)^2}\\[0.2cm] (1-0)^2+(2-k)^2+(0-0)^2&=(2-0)^2+(3-k)^2+(-1-0)^2 \\[0.2cm] 1+4+k^2-4k&=4+9+k^2-6k+1\\[0.2cm] -4k+5&=-6k+14\\[0.2cm] 2k&=9\\[0.2cm] k&= \dfrac{9}{2} \end{aligned}\)

    Answer: k = \( \dfrac{9}{2}\)

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