# Equation of Plane Passing Through Intersection of Two Planes

Here you will learn what is the equation of plane passing through intersection of two planes with examples.

Let’s begin –

## Equation of Plane Passing Through Intersection of Two Planes

### (a) Vector Form

The equation of a plane passing through the intersection of the planes $$\vec{r}.\vec{n_1}$$ = $$d_1$$  and $$\vec{r}.\vec{n_2}$$ = $$d_2$$ is given by

($$\vec{r}.\vec{n_1}$$ – $$d_1$$) + $$\lambda$$($$\vec{r}.\vec{n_2}$$ – $$d_2$$) = 0

or,  $$\vec{r}$$.($$\vec{n_1} + \lambda\vec{n_2}$$) = $$d_1 + \lambda d_2$$,

where $$\lambda$$ is an arbitrary constant.

Example : Find the equation of the plane containing the line of intersection of the planes $$\vec{r}$$.($$\hat{i} + 3\hat{j} – \hat{k}$$) = 5 and $$\vec{r}$$.($$2\hat{i} – \hat{j} + \hat{k}$$) = 3 and passing through the point (2, 1, -2),

Solution : The equation of the plane through the line of intersection of the given planes is

[$$\vec{r}$$.($$\hat{i} + 3\hat{j} – \hat{k}$$) – 5] + $$\lambda$$[$$\vec{r}$$.($$2\hat{i} – \hat{j} + \hat{k}$$) – 3] = 0

$$\implies$$ $$\vec{r}$$.[$$1 + 2\lambda)\hat{i} + (3 – \lambda)\hat{j} + (-1 + \lambda)\hat{k}$$] – 5 – 3$$\lambda$$ = 0                    ………..(i)

If plane in (i) passes through (2, 1, -2), then the vector $$2\hat{i} + \hat{j} – 2\hat{k}$$ should satisfy it

($$2\hat{i} + \hat{j} – 2\hat{k}$$).[$$1 + 2\lambda)\hat{i} + (3 – \lambda)\hat{j} + (-1 + \lambda)\hat{k}$$] – (5 + 3$$\lambda$$) = 0

$$\implies$$ $$-2\lambda + 2$$ = 0 $$\implies$$ $$\lambda$$ = 1

Putting $$\lambda$$ = 1 in equation (i), we obtain the equation of the required plane as

$$\vec{r}$$.($$3\hat{i} + 2\hat{j} + 0\hat{k}$$) = 8.

### (b) Cartesian Form

The equation of a plane passing through the intersection of planes $$a_1x + b_1y + c_1z + d_1$$ = 0 and $$a_2x + b_2y + c_2z + d_2$$ = 0 is

($$a_1x + b_1y + c_1z + d_1$$) + $$\lambda$$($$a_2x + b_2y + c_2z + d_2$$) = 0,

where $$\lambda$$ is a constant.

Example : Find the equation of the plane containing the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through the point (1, 1, 1),

Solution : The equation of the plane through the line of intersection of the given planes is

(x + y + z – 6) + $$\lambda$$(2x + 3y + 4z + 5) = 0                            ………..(i)

If (i) passes through (1, 1, 1), then

-3 + 14 $$\lambda$$ = 0 $$\implies$$ $$\lambda$$ = 3/14

Putting $$\lambda$$ = 3/14 in equation (i), we obtain the equation of the required plane as

(x + y + z – 6) + $$3\over 14$$ (2x + 3y + 4z + 5) = 0

or, 20x + 23y + 26z – 69 = 0.