Here you will learn distance formula in 3d to calculate distance between two points with example.
Let’s begin –
Distance Formula in 3d
The distance between the points P\((x_1, y_1, z_1)\) and Q\((x_2, y_2, z_2)\) is given by
PQ = \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\)
Proof : Let O be the origin and let P\((x_1, y_1, z_1)\) and Q\((x_2, y_2, z_2)\) be two given points. Then,
\(\vec{OP}\) = \(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\), \(\vec{OQ}\) = \(x_2\hat{i} + y_2\hat{j} + z_2\hat{k}\)
Now,
\(\vec{PQ}\) = Position vector Q – Position vector of P
\(\implies\) \(\vec{PQ}\) = \(x_2\hat{i} + y_2\hat{j} + z_2\hat{k}\) – \(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\)
\(\implies\) \(\vec{PQ}\) = \((x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} + (z_2 – z_1)\hat{k}\)
\(\therefore\) PQ = |\(\vec{PQ}\)| = \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\)
Hence, PQ = \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\).
Example : Find the distance between the points P (-2, 4, 1) and Q (1, 2, -5).
Solution : We have, P (-2, 4, 1) and Q (1, 2, -5).
Distance PQ = \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\)
\(\implies\) PQ = \(\sqrt{(1 – (-2))^2 + (2 – 4)^2 + (-5 – 1)^2}\)
\(\implies\) PQ = \(\sqrt{9 + 4 + 36}\) = 7 units
Example : Prove by using the distance formula that the points P(1, 2, 3), Q(-1, -1, -1) and R(3, 5, 7) are collinear.
Solution : We have, P(1, 2, 3), Q(-1, -1, -1) and R(3, 5, 7)
Distance Formula = \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\)
PQ = \(\sqrt{(-1 – 1)^2 + (-2 – 2)^2 + (-1 – 3)^2}\) = \(\sqrt{4 + 9 + 16}\) = \(\sqrt{29}\) units
QR = \(\sqrt{(3 + 1)^2 + (5 + 1)^2 + (7 + 1)^2}\) = \(\sqrt{16 + 36 + 64}\) = \(\sqrt{116}\) units
and, PR = \(\sqrt{(3 – 1)^2 + (5 – 2)^2 + (7 – 3)^2}\) = \(\sqrt{4 + 9 + 16}\) = \(\sqrt{29}\) units
Clearly, QR = PQ + PR.
Therefore, Points P, Q and R are collinear.
