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Family of Plane - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Family of Plane is considered one of the most asked concept.

  • 26 Questions around this concept.

Solve by difficulty

The equation of the plane containing the line 2x-5y+z=3;x+y+4z=5,

and parallel to the plane, x+3y+6z=1,\, is:

Let the equation of the plane passing through the line of intersection of the planes \mathrm{x+2y+az=2 \: and\: x-y+z=3} be \mathrm{5x-11y + bz = 6a -1}. For \mathrm{c\: \epsilon \: \: Z}, if the distance of this plane from the point \mathrm{ (a, -c, c) } is \frac{2}{\sqrt{a}} , then \frac{a+b}{c} is equal

Concepts Covered - 1

Family of Plane

Equation of plane parallel to a given plane

Parallel planes have the same normal vectors, so the equation of plane parallel to \vec {\mathbf r}\cdot \vec {\mathbf n}=d_1 is of the form \vec {\mathbf r}\cdot \vec {\mathbf n}=d_2, where d2 can be determined by using the given conditions.

In cartesian form, if ax + by +cz + d = 0 is given plane, then  the plane parallel to this plane is ax + by + cz + k = 0, where k is any scalar.

 

Plane passing through the intersection of two given planes

Vector Form

Equation of plane passing through the intersection of planes \vec {\mathbf r}\cdot \vec {\mathbf n}_1=d_1 and \vec {\mathbf r}\cdot \vec {\mathbf n}_2=d_2 is 

\vec{\mathbf r} \cdot\left(\vec{\mathbf n}_{1}+\lambda \vec{\mathbf n}_{2}\right)=d_{1}+\lambda d_{2}

Proof:

Let π1 and π2 be two planes with equations \vec {\mathbf r}\cdot \vec {\mathbf n}_1=d_1 and \vec {\mathbf r}\cdot \vec {\mathbf n}_2=d_2 respectively. The position vector of any point on the line of intersection must satisfy both the equations.

If \vec{\mathbf p} is the position vector of a point P on the line, then 
\vec{p} \cdot \hat{n}_{1}=d_{1} \text { and } \vec{p} \cdot \hat{n}_{2}=d_{2}

Therefore, for all real values of λ, we have

\vec{p} \cdot\left(\hat{n}_{1}+\lambda \hat{n}_{2}\right)=d_{1}+\lambda d_{2}

Since, \vec{\mathbf p} is arbitrary, it satisfies for any point on the line.

Hence, the equation \vec{r} \cdot\left(\vec{n}_{1}+\lambda \vec{n}_{2}\right)=d_{1}+\lambda d_{2} represents a plane π3 which is such that if any vector \vec{\mathbf r} satisfies both the equations π1 and π2 , it also satisfies the equation π3 i.e. any plane passing through the intersection of the planes.
 

Cartesian Form

In Cartesian system, let

\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{n}_{1} =\mathrm{a}_{1} \hat{i}+\mathrm{b}_{2} \hat{j}+\mathrm{c}_{1} \hat{k} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \vec{n}_{2} =\mathrm{a}_{2} \hat{i}+\mathrm{b}_{2} \hat{j}+\mathrm{c}_{2} \hat{k} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{\;\;} \vec{r} =x \hat{i}+y \hat{j}+z \hat{k} \\\text{then the vector equation,}\;\;\vec{r} \cdot\left(\vec{n}_{1}+\lambda \vec{n}_{2}\right)=d_{1}+\lambda d_{2}\;\;\text{become}\\\mathrm{\;\;\;\;\;\;\;\;}x\left(\mathrm{a}_{1}+\lambda \mathrm{a}_{2}\right)+y\left(\mathrm{b}_{1}+\lambda \mathrm{b}_{2}\right)+z\left(\mathrm{c}_{1}+\lambda \mathrm{c}_{2}\right)=d_{1}+\lambda d_{2}\\\text{or}\;\;\left(\mathbf{a}_{1} x+\mathbf{b}_{1} y+\mathbf{c}_{1} z-d_{1}\right)+\lambda\left(\mathbf{a}_{2} x+\mathbf{b}_{2} y+\mathbf{c}_{2} z-d_{2}\right)=0

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

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