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.

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