RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.5
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.