# 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 = $$|2\times 2 + 1 + 2\times 0 + 5|\over \sqrt{4 + 1 + 4}$$ = $$10\over 3$$