Here you will learn equation of plane parallel to plane with examples.
Let’s begin –
Equation of Plane Parallel to Plane
(a) Vector Form
Since parallel planes have the common normal, therefore equation of a plane parallel to the plane \(\vec{r}\).\(\vec{n}\) = \(\vec{d_1}\) is
\(\vec{r}\).\(\vec{n}\) = \(\vec{d_2}\)
where \(\vec{d_2}\) is constant determined by the given condition.
Example : Find the equation of plane passing through the point \(\hat{i} + \hat{j} + \hat{k}\) and parallel to the plane \(\vec{r}\).\(2\hat{i} – \hat{j} + 2\hat{k}\) = 5.
Solution : The equation of a plane parallal to the plane \(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = 5 is
\(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = d ………(i)
If it passes through \(\hat{i} + \hat{j} + \hat{k}\), then
(\(\hat{i} + \hat{j} + \hat{k}\)).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = d \(\implies\) 2 – 1 + 2 = d \(\implies\) d = 3
Putting d = 3 in (i), we obtain \(\vec{r}\).(\(2\hat{i} – \hat{j} + 2\hat{k}\)) = 3 as the equation of the required plane.
(b) Cartesian Form
Let ax + by + cz + d = 0 be the cartesian equation of a plane. Then, direction ratios of its normal are proportional to a, b, c.
Since parallel planes have common normals.
Therefore, the direction ratios of the normal to the parallel plane are also proportional to a, b, c.
Thus, the equation of plane parallal to the plane ax + by + cz + d = 0 is
ax + by + cz + k = 0,
where k is an arbitrary constant and is determined by the given condition.
Example : Find the equation of plane passing through the point (1, 4, -2) and parallal to the plane -2x + y – 3z = 7.
Solution : Let the equation of a plane parallal to the plane -2x + y – 3z = 7 be
-2x + y – 3z + k = 7 ………(i)
If it passes through (1, 4, -2), then
(-2)(1) + 4 – 3(-2) + k = 0 \(\implies\) k = -8
Putting k = -8 in (i), we obtain -2x + y – 3z – 8 = 0 as the equation of the required plane.
