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.