Class 12 RD Sharma Mathematics Solutions Chapter 29 The Plane Exercise 29.5 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 29 of RD Sharma's Class 12 Mathematics textbook delves into the concept of planes in three-dimensional space. Exercise 29.5 focuses on problems related to the distance between a point and a plane. This set of solutions demonstrates how to calculate these distances using various formulas and techniques, providing students with practical applications of plane geometry and vector algebra.

**Distance formula between a point and a plane:

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

**where (x₀, y₀, z₀) is the point and Ax + By + Cz + D = 0 is the equation of the plane

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

**Normal vector of a plane: n = Ai + Bj + Ck

Class 12 RD Sharma Mathematics Solutions - Exercise 29.5

**Question 1. Find the vector equation of the plane passing through the points (1, 1, 1), (1, -1, 1) and (-7, -3, -5)

**Solution:

Given that, plane is passing through

(1, 1, 1), (1, -1, 1) and (-7, -3, -5)

We know that, equation of plane passing through 3 points,

\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0

\begin{vmatrix}x-1 & y-1 & z-1\\ 1-1 & -1-1 & 1-1\\ -7-1 & -3-1 & -5-1\end{vmatrix}=0

\begin{vmatrix}x-1 & y-1 & z-1\\ 0 & -2 & 0\\ -8 & -4 & -6\end{vmatrix}=0

(x - 1)(12 - 0) - (y - 1)(0 - 0) + (z - 1)(0 - 16) = 0

(x - 1)(12) - (y - 1)(0) + (z - 1)(-16) = 0

12x - 12 - 0 - 16z + 16 = 0

12x - 16z + 4 = 0

Dividing by 4,

3x - 4z + 1 = 0

(x\hat{i}+y\hat{j}+z\hat{k})(3\hat{i}+0\hat{j}-4\hat{k})+1=0\\ \vec{r}.(3\hat{i}-4\hat{k})+1=0

Equation of the required plane,

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

**Question 2. Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).

**Solution:

Let P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3) be the three points on a plane having position vectors \vec{p},\vec{q}\ and\ \vec{s} respectively. Then the vectors \overrightarrow{PQ} and \overrightarrow{PR} are in the same plane. Therefore, \overrightarrow{PQ}\times\overrightarrow{PR} is a vector perpendicular to the plane.

Let = \vec{n} = \overrightarrow{PQ}\times\overrightarrow{PR}

\overrightarrow{PQ}=(-2-2)\hat{i}+(-3-5)\hat{j}+(5-(-3))\hat{k}\\ \overrightarrow{PQ}=-4\hat{i}-8\hat{j}+8\hat{k}

Similarly,

\overrightarrow{PR}=(5-2)\hat{i}+(3-5)\hat{j}+(-3-(-3))\hat{k}\\ \overrightarrow{PQ}=3\hat{i}-2\hat{j}+0\hat{k}

Thus

\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}\\ =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ -4 & -8 & 8\\ 3 & -2 & 0\end{vmatrix}\\ =16\hat{i}+24\hat{j}+32\hat{k}

The plane passes through the point P with position vector \vec{p}=2\hat{i}+5\hat{j}-3\hat{k}

Thus, its vector equation is

(\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\ (\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-(32+120-96)=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-56=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})=56\\ ⇒\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7

**Question 3. Find the vector equation of the plane passing through the points A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from origin, prove that \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}

**Solution:

Let A(a, 0, 0), B(0, b, 0) and C(0, 0, c) be three points on a plane having their position vectors \vec{a},\vec{b}\ and\ \vec{c} respectively. Then vectors \overrightarrow{AB} and \overrightarrow{AC} are in the same plane. Therefore, \overrightarrow{AB}\times\overrightarrow{AC} is a vector perpendicular to the plane.

Let \vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}

\overrightarrow{AB}=(0-a)\hat{i}+(b-0)\hat{j}+(0-0)\hat{k}\\ \overrightarrow{AB}=-a\hat{i}+b\hat{j}+0\hat{k}

Similarly,

\overrightarrow{AC}=(0-a)\hat{i}+(0-0)\hat{j}+(c-0)\hat{k}\\ \overrightarrow{AC}=-a\hat{i}+0\hat{j}+c\hat{k}

Thus

\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}

\hat{i}\ \ \ \hat{j}\ \ \ \hat{k}

= | -a b 0 |

-a 0 c

\vec{n}=bc\hat{i}+ac\hat{j}+ab\hat{k}

⇒\hat{n}=\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}

The plane passes through the point P with position vector \vec{a}=a\hat{i}+0\hat{j}+0\hat{k}

Thus, the vector equation in the normal form is

\{\vec{r}-\left(a\hat{i}+0\hat{j}+0\hat{k}\right)\}.\left(\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\right)=0\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}...(1)

The vector equation of a plane normal to the unit vector \hat{n} and at a distance 'd' from the origin is \vec{r}.\hat{n}=d ....(2).

Given that the plane is at a distance 'p' from the origin.

Comparing equations (1) and (2), we have,

d = p = \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\\ ⇒\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}

**Question 4. Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).

**Solution:

Let P(1, 1, -1), Q(6, 4, -5) and R(-4, -2, 3) be three points on a plane having position vectors \vec{p},\ \vec{q}\ and\ \vec{s} respectively. Then the vectors \overrightarrow{PQ}\ and\ \overrightarrow{PR} are in the same plane. Therefore, \overrightarrow{PQ}\times\overrightarrow{PR} is a vector perpendicular to the plane.

Let \vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}

\overrightarrow{PQ}=(6-1)\hat{i}+(4-1)\hat{j}+(-5-(-1))\hat{k}\\ \overrightarrow{PQ}=5\hat{i}+3\hat{j}-4\hat{k}

Similarly,

\overrightarrow{PR}=(-4-1)\hat{i}+(-2-1)\hat{j}+(3-(-1))\hat{k}\\ \overrightarrow{PR}=-5\hat{i}-3\hat{j}+4\hat{k}

Thus

Here, \overrightarrow{PQ}=-\overrightarrow{PR}

Therefore, the given points are collinear.

Thus, \vec{n}=a\hat{i}+b\hat{j}+c\hat{k} where, 5a + 3b - 4c = 0

The plane passes through the point P with position vector \vec{p}=\hat{i}+\hat{j}-\hat{k}

Thus, its vector equation is

\{\vec{r}-(\hat{i}+\hat{j}-\hat{k})\}.(a\hat{i}+b\hat{j}+c\hat{k})=0 , where, 5a + 3b - 4c = 0

**Question 5. Find the vector equation of the plane passing through the points

3\hat{i}+4\hat{j}+2\hat{k},\ 2\hat{i}-2\hat{j}-\hat{k}\ \ and\ \ 7\hat{i}+6\hat{k}

**Solution:

Let A, B, C be the points with position vector (3\hat{i}+4\hat{j}+2\hat{k}),(2\hat{i}-2\hat{j}-\hat{k})\ and\ (7\hat{i}+6\hat{k})

respectively. Then

\overrightarrow{AB} = Position vector of B - Position vector of A

=(2\hat{i}-2\hat{j}-\hat{k})-(3\hat{i}+4\hat{j}+2\hat{k})\\ =2\hat{i}-2\hat{j}-\hat{k}-3\hat{i}-4\hat{j}-2\hat{k}\\ =-\hat{i}-6\hat{j}-3\hat{k}

\overrightarrow{BC} = Position vector of C - Position vector of B

=(7\hat{i}+6\hat{k})-(2\hat{i}-2\hat{j}-\hat{k})\\ =7\hat{i}+6\hat{k}-2\hat{i}+2\hat{j}+\hat{k}\\ =5\hat{i}+2\hat{j}+7\hat{k}

A vector normal to A, B, C is a vector perpendicular to \overrightarrow{AB}\times\overrightarrow{BC}

\vec{n}=\overrightarrow{AB}\times\overrightarrow{BC}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -1&-6&-3\\ 5&2&7\end{vmatrix}\\ \vec{n}=\hat{i}(-42+6)-\hat{j}(-7+15)+\hat{k}(-2+30)\\ =-36\hat{i}-8\hat{j}+28\hat{k}

As we know that, equation of a plane passing through vector \vec{a} and perpendicular to vector \vec{n} is given by,

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

Put \vec{a} and \vec{n} in equation (1)

\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k})=(3\hat{i}+4\hat{j}+2\hat{k})(-36\hat{i}-8\hat{j}+28\hat{k})

= (3)(-36) + (4)(-8) + (2)(28)

= -108 - 32 + 56

= -140 + 56

\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k}) = -84

Dividing by (-4), we will get

\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21

Equation of required plane is,

\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21

Practice Questions:

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

2. Determine the angle between the planes 3x + 2y - 6z + 1 = 0 and x - y + z - 2 = 0.

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

4. Calculate the angle between the planes x + y + z = 1 and 2x - y + 2z = 0.

5. Determine if the planes 3x - 6y + 2z = 7 and 9x - 18y + 6z = 5 are perpendicular.

6. Find the angle between the planes x + 2y + 2z = 0 and 2x + 4y - z = 0.

7. Calculate the acute angle between the planes 2x + 2y + z = 0 and x - y + 2z = 0.

8. Determine if the planes 2x + 3y - 6z = 5 and 4x + 6y + 3z = 2 are perpendicular.

9. Find the angle between the planes 3x - 4y + 12z = 2 and 6x + 8y - 3z = 1.

10. Calculate the angle between the planes ax + by + cz = 0 and bx + cy + az = 0, where a, b, and c are non-zero real numbers.

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Conclusion

Exercise 29.5 provides valuable practice in applying the distance formula between points and planes, as well as working with perpendiculars to planes. These problems demonstrate the interplay between algebraic and geometric approaches in three-dimensional space. Mastering these concepts is crucial for students as they form the foundation for more advanced topics in multivariable calculus and physics. The skills developed here, such as manipulating vector equations and visualizing spatial relationships, are essential for many fields of study, including engineering, computer graphics, and theoretical physics.