Distance of a Point From a Plane

Here you will learn how to find the distance of a point from a plane formula with examples.

Let’s begin –

Distance of a Point from a Plane

(a) Vector Form

The length of the perpendicular from a point having position vector \(\vec{a}\) to the plane \(\vec{r}\).\(\vec{n}\) = d  is

p = \(|\vec{a}.\vec{n} – d|\over |\vec{n}|\)

(b) Cartesian Form

The length of the perpendicular form a point P\((x_1, y_1, z_1)\) to the plane ax + by + cz + d = 0 is given by

p = \(|ax_1 + by_1 + cz_1 + d|\over \sqrt{a^2 + b^2 + c^2}\)

Example : Find the distance of the point \(2\hat{i} + \hat{j} – \hat{k}\) from the plane \(\vec{r}\).\(\hat{i}  – 2\hat{j} + 4\hat{k}\) = 9.

Solution : We know that the perpendicular distance of a point with the position vector \(\vec{a}\) from the given planes \(\vec{r}\).\(\vec{n}\) = d is given by

p = \(|\vec{a}.\vec{n} – d|\over |\vec{n}|\)

Here, \(\vec{a}\) = \(2\hat{i} + \hat{j} – \hat{k}\)  ,  \(\vec{n}\) = \(\hat{i}  – 2\hat{j} + 4\hat{k}\) and d = 9.

So, the required distance p is given by

p = \(|(2\hat{i} + \hat{j} – \hat{k}).(\hat{i}  – 2\hat{j} + 4\hat{k}) – 9|\over \sqrt{1 + 4 + 16}\)

= \(|2 – 2 – 4 – 9|\over \sqrt{21}\) = \(13\over \sqrt{21}\)

Example : Find the distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0.

Solution : We know that the perpendicular distance of the point \((x_1, y_1, z_1)\) from the given planes ax+ by + cz + d = 0 is given by

p = \(|ax_1 + by_1 + cz_1 + d|\over \sqrt{a^2 + b^2 + c^2}\)

So, the required distance p is

p = \(|2\times 2 + 1 + 2\times 0 + 5|\over \sqrt{4 + 1 + 4}\) = \(10\over 3\)