Class 12 RD Sharma Solutions Chapter 29 The Plane Exercise 29.7 (original) (raw)

Last Updated : 23 Jul, 2025

**Chapter 29 of RD Sharma's Class 12 Mathematics textbook focuses on the study of planes in three-dimensional space. Exercise 29.7 delves into more advanced concepts related to planes, including finding the equations of planes under various conditions, determining the angles between planes, and solving problems involving multiple planes. This exercise builds upon the fundamental concepts introduced earlier in the chapter and challenges students to apply their knowledge to more complex scenarios.

**General equation of a plane: Ax + By + Cz + D = 0

**Angle between two planes: cos θ = |A₁A₂ + B₁B₂ + C₁C₂| / √((A₁² + B₁² + C₁²)(A₂² + B₂² + C₂²))

**Condition for perpendicular planes: A₁A₂ + B₁B₂ + C₁C₂ = 0

**Distance of a point (x₀, y₀, z₀) from a plane Ax + By + Cz + D = 0:

**d = |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)

Class 12 RD Sharma Mathematics Solutions - Exercise 29.7

**Question 1. Find the vector equation of the following planes in scalar product form (\vec{r}.\vec{n}=d):

**Solution:

****(i)** \vec{r}=\{2\hat{i}-\hat{k}\}+λ \hat{i}+µ\{\hat{i}-2\hat{j}-\hat{k}\}

Here, \vec{r}=\{2\hat{i}-\hat{k}\}+λ \hat{i}+µ\{\hat{i}-2\hat{j}-\hat{k}\}

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represent a plane passing through a point having position vector \vec{a} and parallel to vectors \vec{b} and .\vec{c}

Here, \vec{a}=2\hat{i}-\hat{k},\ \vec{b}=\hat{i},\ \vec{c}=\hat{i}-2\hat{j}-\hat{k}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&0\\1&-2&-1\end{vmatrix}\\ =\hat{i}(0-0)-\hat{j}(-1-0)+\hat{k}(-2-0)\\ =0\hat{i}+\hat{j}+2\hat{k}\\ \vec{n}=\hat{j}-2\hat{k}

We know that vector equation of plane in scalar product form is,

\vec{r}.\vec{n}=\vec{a}.\vec{n} ---(Equation-1)

Put \vec{n} and \vec{a} in (Equation-1),

\vec{r}.(\hat{j}-2\hat{k})=(2\hat{i}-\hat{k})(\hat{j}-2\hat{k})\\ \vec{r}.(\hat{j}-2\hat{k})=(2)(0)+(0)(1)+(-1)(-2)\\ =0+0+2\\ \vec{r}.(\hat{j}-2\hat{k})=2

The equation is required form is,

\vec{r}.(\hat{j}-2\hat{k})=2

****(ii)** \vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}

Here,\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}\\ \vec{r}=(1+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(-\hat{i}+2\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represent a plane passing through a point having position vector \vec{a} and parallel to vectors \vec{b} and \vec{c}

Here, \vec{a}=\hat{i}+2\hat{j}+3\hat{k},\ \vec{b}=\hat{i}-\hat{j}-2\hat{k},\ \vec{c}=-\hat{i}+2\hat{k}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&-2\\-1&0&2\end{vmatrix}\\ =\hat{i}(-2-0)-\hat{j}(2-2)+\hat{k}(0-1)\\ \vec{n}=-2\hat{i}-\hat{k}

We know that, vector equation of a plane is scalar product is,

\vec{r}.\vec{n}=\vec{a}.\vec{n} ---(Equation-1)

Put value of \vec{a} and \vec{n} in (Equation-1)

\vec{r}.(-2\hat{i}-\hat{k})=(-2\hat{i}-k)(\hat{i}+2\hat{k}+3\hat{k})\\ \vec{r}.(-2\hat{i}-\hat{k})=(-2)(1)+(0)(2)+(-1)(3)\\ =-2+0-3\\ \vec{r}.(-2\hat{i}-\hat{k})=-5

Multiplying both the sides by (-1),

\vec{r}.(2\hat{i}+\hat{k})=5

The equation in the required form,

\vec{r}.(2\hat{i}+\hat{k})=5

****(iii)** \vec{r}=(\hat{i}+\hat{j})+λ(\hat{i}+2\hat{j}-\hat{k})+μ(-\hat{i}+\hat{j}-2\hat{k})

Given, equation of plane,

\vec{r}=(\hat{i}+\hat{j})+λ(\hat{i}+2\hat{j}-\hat{k})+μ(-\hat{i}+\hat{j}-2\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} is the equation of a plane passing through point \vec{a} and parallel to \vec{b} and \vec{c} .

Here, \vec{a}=\hat{i}+\hat{j},\ \vec{b}=\hat{i}+2\hat{j}-\hat{k},\ \vec{c}=-\hat{i}+\hat{j}-2\hat{k}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-1\\-1&1&-2\end{vmatrix}\\ =\hat{i}(-4+1)-\hat{j}(-2-1)+\hat{k}(1+2)\\ -3\hat{i}+3\hat{j}+3\hat{k}

We know that, equation of plane in scalar product form is given by,

\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=(\hat{i}+\hat{j})(-3\vec{i}+3\vec{j}+3\hat{k})\\ =(1)(-3)+(1)(3)+(0)(3)\\ =-3+3\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=0

Dividing by 3, we get

\vec{r}.(-\vec{i}+\vec{j}+\hat{k})=0

Equation in required form is,

\vec{r}.(-\vec{i}+\vec{j}+\hat{k})=0

****(iv)** \vec{r}=\hat{i}-\hat{j}+λ(\hat{i}+\hat{j}+\hat{k})+μ(4\hat{i}-2\hat{j}-3\hat{k})

\vec{r}=\hat{i}-\hat{j}+λ(\hat{i}+\hat{j}+\hat{k})+μ(4\hat{i}-2\hat{j}-3\hat{k})

Plane is passing through (\hat{i}-\hat{j}) and parallel to b b(\hat{i}+\hat{j}+\hat{k}) and c(4\hat{i}-2\hat{j}+3\hat{k})

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\4&-2&3\end{vmatrix}\\ n=5i+j-6k\\ \vec{r}.n=(i-j)(5i+j-6k)=5-1=4\\ \vec{r}.(5i+j-6k)=4

**Question 2. Find the cartesian form of the equation of the following planes:

**Solution:

****(i)** \vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{i}+\hat{k})\\

Here, given equation of plane is,

\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{i}+\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represents the equation of a plane passing through a vector \vec{a} and parallel to vector \vec{b} and \vec{c} .

Here, \vec{a}=\hat{i}-\hat{j},\ \vec{b}=-\hat{i}+\hat{j}+2\hat{k},\ \vec{c}=\hat{i}+2\hat{j}+\hat{k}

Given plane is perpendicular to vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&2\\1&2&1\end{vmatrix}\\ =\hat{i}(1-4)-\hat{j}(-1-2)+\hat{k}(-2-1)\\ \vec{n}=-3\hat{i}+3\hat{j}-3\hat{k}

We know that, equation of plane in the scalar product form,

\vec{r}.\vec{n}=\vec{a}.\vec{n} ---Equation-1

Put the value of \vec{a} and \vec{n} in Equation-1,

\vec{r}.(\hat{i}-\hat{j})=(\hat{i}-\hat{j})(-3\hat{i}+3\hat{j}-3\hat{k})\\ \vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=(1)(-3)+(-1)(3)+(0)(-3)\\ =-3-3+0\\ \vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=-6

Put \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ (x\hat{i}+y\hat{j}+z\hat{k})(-3\hat{i}+3\hat{j}-3\hat{k})=-6

(x)(-3) + (y)(3) + (z)(-3) = -6

-3x + 3y - 3z = -6

Dividing by (-3), we get

x - y + z = 2

Equation in required form is,

x - y + z = 2

****(ii)** \vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}

Given, equation of plane,

\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}\\ =(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(\hat{i}+\hat{j}+2\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represents the equation of a plane passing through the vector \vec{a} and parallel to vector \vec{b} and \vec{c}

Here, \vec{a}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{b}=\hat{i}-\hat{j}-2\hat{k}\\ \vec{c}=\hat{i}+\hat{j}+2\hat{k}

The given plane is perpendicular to vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&-2\\1&1&2\end{vmatrix}\\ =\hat{i}(-2+2)-\hat{j}(2+2)+\hat{k}(1+1)\\ =0(\hat{i})-4\hat{j}+2\hat{k}\\ \vec{n}=-4\hat{j}+2\hat{k}

We know that, equation of plane in scalar product form is given by,

\vec{r}.\vec{n}=\vec{a}.\vec{n} ---Equation-1

Put, the value of \vec{a} and \vec{n} in equation-1

\vec{r}.(-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k})(-4\hat{j}+2\hat{k})\\ \vec{r}.(-4\hat{j}+2\hat{k})=(1)(0)+(2)(-4)+(3)(2)\\ =0-8+6\\ \vec{r}.(-4\hat{j}+2\hat{k})=-2

Put \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

(x\hat{i}+y\hat{j}+z\hat{k})(-4\hat{j}+2\hat{k})=-2\\

(x)(0) + (y)(-4) + (z)(2) = -2

-4y + 2z = -2

The equation in required form is,

2y - z = 1

**Question 3. Find the vector equation of the following planes in non-parametric form:

**Solution:

****(i)** \vec{r}=(λ-2μ)\hat{i}+(3-μ)\hat{j}+(2λ+μ)\hat{k}

Given, equation of plane is,

\vec{r}=(λ-2μ)\hat{i}+(3-μ)\hat{j}+(2λ+μ)\hat{k}\\ \vec{r}=(3\hat{j})+λ(\hat{i}+2\hat{k})+μ(-2\hat{i}-\hat{j}+\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represents the equation of a plane passing through a point \vec{a} and parallel to vector \vec{b} and \vec{c} .

Given,

\vec{a}=3\hat{j}\\ \vec{b}=\hat{i}+2\hat{k}\\ \vec{c}=-2\hat{i}-\hat{j}+\hat{k}

The given plane is perpendicular to

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&2\\-2&-1&1\end{vmatrix}\\ =\hat{i}(0+2)-\hat{j}(1+4)+\hat{k}(1-0)\\ \vec{n}=2\hat{i}-5\hat{j}-\hat{k}

Vector equation of plane in non-parametric form is.

\vec{r}.\vec{n}=\vec{a}\vec{n}\\ \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=(3\hat{i})(2\hat{i}-5\hat{j}-\vec{k})

= (0)(2) + (3)(-5) + (0)(-1)

= 0 - 15 + 0

\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=-15\\ \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0

The required form of equation is,

\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0

****(ii)** \vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\

Given, equation of plane is,

\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c} represents the equation of a plane passing through a vector \vec{a} and parallel to vector \vec{b} and \vec{c} .

Here,

\vec{a}=2\hat{i}+2\hat{j}-\hat{k}\\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{c}=5\hat{i}-2\hat{j}+7\hat{k}

The given plane is perpendicular to vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\5&-2&7\end{vmatrix}\\ =\hat{i}(14+6)-\hat{j}(7-15)+\hat{k}(-2-10)\\ \vec{n}=20\hat{i}+8\hat{j}-12\hat{k}

We know that, equation of a plane in non-parametric form is given by,

\vec{r}.\vec{n}=\vec{a}.\vec{n}

\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=(2\hat{i}+2\hat{j}-\hat{k})(20\hat{i}+8\hat{j}-12\hat{k})\\

= (2)(20) + (2)(8) - (-1)(-12)

=40 + 16 + 12

\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=68

Dividing by 4,

\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17

Equation of plane in required form is,

\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17

Practice Questions:

**Question 1. Find the equation of the plane passing through the point (1, 2, 3) and perpendicular to the vector <2, -1, 4>.

**Question 2. Determine the equation of the plane that passes through the points (0, 0, 0), (1, 2, 3), and (2, 1, -1).

**Question 3. Find the distance of the point (2, 3, 4) from the plane 2x + 3y - z + 5 = 0.

**Question 4. Write the equation of the plane passing through the point (1, -1, 2) and parallel to the plane 3x - 2y + z = 5.

**Question 5. Find the angle between the planes 2x + y - 2z + 3 = 0 and x - 3y + z - 4 = 0.

**Question 6. Determine if the planes 2x + 3y - z = 4 and 4x + 6y - 2z = 8 are parallel.

**Question 7. Find the equation of the plane passing through the line of intersection of the planes x + y + z = 1 and 2x - y + z = 3, and perpendicular to the plane x - y + 2z = 4.

**Question 8. Calculate the shortest distance between the parallel planes 2x + 2y - z = 5 and 2x + 2y - z = 9.

**Question 9. Find the equation of the plane which is at a distance of 3 units from the origin and normal to the line x/2 = y/3 = z/4.

**Question 10. Determine if the point (1, 2, 3) lies on the plane 2x - y + 2z = 9.

**Also Read,

Conclusion

Exercise 29.7 of RD Sharma's Class 12 Chapter 29 presents a set of challenging problems that require a deep understanding of planes in three-dimensional space. These solutions demonstrate various techniques for working with planes, including finding equations of planes under specific conditions, determining angles between planes, and solving problems involving multiple planes. By mastering these concepts, students develop advanced skills in analytical geometry and vector algebra, which are crucial for further studies in mathematics, physics, and engineering. The problems in this exercise emphasize the importance of spatial reasoning and the application of vector operations in solving complex geometric scenarios.