RD Sharma Class 12 Exercise 28.3 The Plane Solutions Maths

RD Sharma Class 12 Exercise 28.3 The Plane Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.3

The Plane exercise 28.3 question 1

Answer:
$\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=3$
Hint:
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Given:
Plane passing through the point having the position vector $2 \hat{\imath}-\hat{\jmath}+\hat{k}$ and perpendicular to vector $4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}$
Solution:
$\begin{aligned} &\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ &\vec{n}=4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k} \end{aligned}$
Vector equation of the plane passing through point a and normal to n is ,
$\begin{aligned} &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \\ &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \end{aligned}$
$\begin{aligned} &\Rightarrow \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \\ &=\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=(8-2-3)=3 \\ &=\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=3 \end{aligned}$

The Plane exercise 28.3 question 2(i)

Answer:
$12x-3y+4z+5=0$
Hint:
$\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
Given:
$\vec{r} \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+5=0$
Solution:
$\begin{aligned} &\vec{r} \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+5=0 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+5=0 \\\\ &\Rightarrow 12 x-3 y+4 z+5=0 \end{aligned}$

The Plane exercise 28.3 question 2(ii)

Answer:
$-x+y+2z=9$
Hint:
$\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
Given:
$\vec{r} \cdot(-\hat{\imath}+\hat{\jmath}+2 \hat{k})=9$
Solution:
$\begin{aligned} &\vec{r} \cdot(-\hat{\imath}+\hat{\jmath}+2 \hat{k})=9 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(-\hat{\imath}+\hat{\jmath}+2 \hat{k})=9 \\\\ &-x+y+2 z=9 \end{aligned}$

The Plane exercise 28.3 question 3

Answer:
$\vec{r} \cdot \hat{\imath}=0, \quad \vec{r}, \hat{\jmath}=0, \quad \vec{r} . \hat{k}=0$
Hint:
xy-plane?z=0,
yz-plane?x=0,
zx-plane?y=0,
Given:
Vector equations of the coordinate plane.
Solution:

Vector equ of XY-plane

∴cartesian equ of xy-plane

$\begin{aligned} &z=0 \\\\ &\Rightarrow 0 x+0 y+z=0 \\\\ &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(0 \hat{\imath}+0 \hat{\jmath}+\hat{k})=0 \\\\ &\vec{r} \cdot \hat{k}=0[\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}] \end{aligned}$

Vector equ of yz-plane

∴cartesian equ of yz-plane

$\begin{aligned} &x=0 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+0 \hat{\jmath}+0 \hat{k})=0 \\\\ &\Rightarrow \vec{r} \cdot \hat{\imath}=0 \end{aligned}$

Vector equ of xz-plane

cartesian equ of xz-plane ∴y=0

$\begin{aligned} &y=0 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(0 \hat{\imath}+\hat{\jmath}+0 \hat{k})=0 \\\\ &\Rightarrow \vec{r} \cdot \hat{\jmath}=0 \end{aligned}$
The vector equation of co-ordinate plane is
$\vec{r} \cdot \hat{\imath}=0, \quad \vec{r} \cdot \hat{\jmath}=0, \quad \vec{r} \cdot \hat{k}=0$

The Plane exercise 28.3 question 4(i)

Answer:
$\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})=8$
Hint:
$\vec{r} \cdot \vec{n}=d$
Given:
$2x-y+2z=8$
Solution:
Vector equation of the plane is $\vec{r} \cdot \vec{n}=d$
$\begin{aligned} &2 x-y+2 z=8 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})=8 \\\\ &\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})=8[\therefore \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}] \end{aligned}$

The Plane exercise 28.3 question 4(ii)

Answer:
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}-\hat{k})=5$
Hint:
$\vec{r} \cdot \vec{n}=d$
Given:
$x+y-z=5$
Solution:
Vector equation of the plane $\vec{r} \cdot \vec{n}=d$
$\begin{aligned} &x+y-z=5 \\\\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}-\hat{k})=5 \\\\ &\vec{r} \cdot(\hat{\imath}+\hat{\jmath}-\hat{k})=5[\therefore \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}] \end{aligned}$

The Plane exercise 28.3 question 4(iii)

Answer:
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath})=3$
Hint:
$\vec{r} \cdot \vec{n}=d$
Given:
$x+y=5$
Solution:
Vector equation of the plane $\vec{r} \cdot \vec{n}=d$
$x+y=3$
$\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath})=3 \\\\ &\vec{r} \cdot(\hat{\imath}+\hat{\jmath})=3[\therefore \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}] \end{aligned}$

The Plane exercise 28.3 question 5

Answer:
$\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7, \quad 2 x-y+4 z=7$
Hint:
Equation of plane, $a(x-1)+b(y+1)+c(z-1)=0$
Given:
Plane passing through the point (1,-1,1) and normal to the line joining the points (1,2,5) and (-1,3,1).
Solution:
Equation of plane, $a(x-1)+b(y+1)+c(z-1)=0$
Where a, b, c are direction ratio
$A = (1,2,5)$ and $B = (-1,3,1)$
$A B=\frac{x}{-2}=\frac{y}{1}=\frac{z}{-4}$
$(-2, 1, -4)$ are direction ratio.
Direction ratio of the plane will be equal to direction ratio of the line joining two point
$\therefore a=-2, b=1, c=-4$
Put the value of a, b, c in equ (1)
Equ of plane
$\begin{aligned} &-2(x-1)+1(y+1)+(-4)(z-1)=0 \\\\ &\Rightarrow-2 x+2+y+1-4 z+4=0 \\\\ &\Rightarrow-2 x+y-4 z=-7 \end{aligned}$
Vector equation
$\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7, \quad[\therefore \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}]$
For cartesian equation, substitute $\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7 \\\\ &2 \mathrm{x}-\mathrm{y}+4 \mathrm{z}=7 \end{aligned}$

The Plane exercise 28.3 question 6

Answer:
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2, \quad x+y+z=2$
Hint:
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Given:
Plane passing through $(2,1,-1)$
Solution:
$|\vec{n}|=\sqrt{3}$
Let $\vec{n}$ vector make $\alpha$ angle with x-axis, β angle with y-axis and $\gamma$ angle with z-axis.
Since
$\begin{aligned} &\alpha=\beta=\gamma \\ &\cos \alpha=\cos \beta=\cos \gamma \\ &l=m=n \end{aligned}$
$\begin{aligned} &\therefore l^{2}+m^{2}+n^{2}=1 \\ &\Rightarrow l^{2}+l^{2}+l^{2}=1 \\ &\Rightarrow 3 l^{2}=1 \\ &\Rightarrow l^{2}=\pm \frac{1}{\sqrt{3}} \end{aligned}$
$\therefore l=\frac{1}{\sqrt{3}}=m=n$ [ since its make acute angle ]
$\begin{aligned} &\hat{n}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k} \\ &\frac{\vec{n}}{|\hat{n}|}=\frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k}) \end{aligned}$
$\begin{aligned} &\vec{n}=\frac{\sqrt{3}}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k}) \ldots|\vec{n}|=\sqrt{3} \\ &\vec{n}=(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \end{aligned}$
$\begin{aligned} &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=(2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2+1-1=2 \end{aligned}$
For cartesian form,substitute $\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
$\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2 \\ &\Rightarrow x+y+z=2 \end{aligned}$

The Plane exercise 28.3 question 7

Answer:
$12x-4y+3z=169$
Hint:
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Given:
The co-ordinates of the foot of the perpendicular drawn from the origin to a plane are
$(12,-4, 3)$
Solution:
Let $O (0,0,0)$ and $P(12, -4, 3)$
$\begin{aligned} &d . r \text { of } O P=12-0, \quad 4-0, \quad 3-0 \\ &=12,-4,3 \end{aligned}$
$\begin{aligned} &\vec{n}=12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k} \\ &\vec{a}=12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k} \\ &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \end{aligned}$
$\begin{aligned} &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k})=(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k}) \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k}) \\ &\Rightarrow \vec{r} \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k})=169 \end{aligned}$ ..is the required vector equation
$\begin{aligned} &\Rightarrow 12 x-4 y+3 z=144+16+9 \\ &\Rightarrow 12 x-4 y+3 z=169 \end{aligned}$ ..is the required equation of plane

The Plane exercise 28.3 question 8

Answer:
$5x+3y+2z=21$
Hint:
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given:
Equation of Plane passing through $(2,3,1)$ & d.r. $(5,3,2)$
Solution:
Equation of Plane passing through $\left(x_{1}, y_{1}, z_{1}\right)$ and DR of the plane proportional to $(a,b,c)$
$\begin{aligned} &a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0 \\ &\text { Here, }\left(x_{1}, y_{1}, z_{1}\right)=(2,3,1) \&(a, b, c)=(5,3,2) \\ &\Rightarrow 5(x-2)+3(y-3)+2(z-1)=0 \\ &\Rightarrow 5 x-10+3 y-9+2 z-2=0 \\ &\Rightarrow 5 x+3 y+2 z=21 \end{aligned}$

The Plane exercise 28.3 question 9

Answer:
The answer of given question is $2x+3y-z=14$
Hint:
By using formula $\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
P is the point $(2,3,-1)$ and the require plane is passing through P at right angle to OP.
Solution:
As per the given criteria, it means that the plane is passing through P and OP is the vector normal to the plane.
Let the position vector of the point P be.
$\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots \ldots(i)$
And it is also given the planes is normal to the line joining the points O(0,0,0) and P(2,-3,1).
Then $\vec{n}=\overrightarrow{O P}$
? $\vec{n}$ =position vector of $\vec{P}$ $-$ Position vector of $\vec{n}$
$\begin{aligned} &\vec{n}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots \ldots .(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equ (i) and equ (ii) in the above equ we get
$\begin{aligned} &{[\vec{r}-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})] \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0 \\ &\Rightarrow \vec{r}-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[(2)(2)+(3)(3)+(-1)(-1)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[4+9+1]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-14=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=14 \end{aligned}$
Then, the above vector equation of the plane becomes
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-3 \hat{k})=14$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(2)+(y)(3)+(z)(-1)=14 \\ &\Rightarrow 2 x+3 y-z=14 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 10

Answer:
The answer of given question is $\frac{2}{3}, \frac{1}{3},-\frac{2}{3}$
Hint:
Dividing by 3 on both sides.
Given:
The given equation of the plane is $2x+y-2z=3$
Solution:
Dividing by 3 on both sides, we get
$\begin{aligned} &\frac{2 x}{3}+\frac{y}{3}-\frac{2 z}{3}=\frac{3}{3} \\ &\Rightarrow \frac{\frac{x}{3}}{2}+\frac{y}{3}-\frac{\frac{z}{3}}{2}=1 \ldots (i) \end{aligned}$
We know that if a,b,c are the intercepts by the plane on thee co-ordinates axes, then equation of the plane is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Comparing the equation (i) and (ii) we get
$a=\frac{3}{2}, \quad b=3, \quad c=-\frac{3}{2}$
Again the given equation of the plane is $2x+y-2z=3$ writing this in vector form, we get
$\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})=3 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})=3 \end{aligned}$
So vector normal to the plane is given by
$\begin{aligned} &\vec{n}=2 \hat{\imath}+\hat{\jmath}-2 k \\ &\Rightarrow|\vec{n}|=\sqrt{(2)^{2}+(1)^{2}+(-2)^{2}} \\ &\Rightarrow|\vec{n}|=\sqrt{4+1+4} \\ &\Rightarrow|\vec{n}|=3 \end{aligned}$
Direction vector of $\vec{n}=2,1,-2$
Direction vector of $\vec{n}$
$\begin{aligned} &=\frac{2}{|\vec{n}|}, \frac{1}{|\vec{n}|}, \frac{-2}{|\vec{n}|} \\ &\Rightarrow \vec{n}=\frac{2}{3}, \frac{1}{3},-\frac{2}{3} \end{aligned}$
So,
Intercepts by the plane on the co-ordinate axes are $=\frac{2}{3}, \frac{1}{3},-\frac{2}{3}$
Direction cosines of normal to the plane are $=\frac{2}{3}, \frac{1}{3},-\frac{2}{3}$

The Plane exercise 28.3 question 11

Answer:
$3x+y-z=-4$ is the cartesian form of equation of the required plane &
$\vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0$ is the vector equation of a required plane
Hint:
Let the position vector of this point Q be $\vec{a}=\hat{\imath}-2 \hat{\jmath}+5 \hat{k} \ldots(i)$

Given:
plane is passing through $(1,-2,5)$ and is perpendicular to OP, where point O is the origin and position vector of point P is $3 \hat{\imath}+\hat{\jmath}-\hat{k}$
Solution:
As per the criteria required plane is passing through $Q(1,-2,5)$ and is perpendicular to OP, where point O is the origin and position vector of point P is $3 \hat{\imath}+\hat{\jmath}-\hat{k}$
Let the position vector of the point Q be.
$\vec{a}=\hat{\imath}-2 \hat{\jmath}+5 \hat{k}$
And it is also given the planes is normal to the line joining the points $O(0,0,0)$ and position vector of point P is $3 \hat{\imath}+\hat{\jmath}-\hat{k}$
Then
$\vec{n}=\overrightarrow{O P}$
? $\vec{n}$ = position vector of $\vec{P}$ $-$ Position vector of $\vec{O}$
$\begin{aligned} &\vec{n}=(3 \hat{\imath}+\hat{\jmath}-\hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\vec{n}=3 \hat{\imath}+\hat{\jmath}-\hat{k} \ldots(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(\hat{\imath}-2 \hat{\jmath}+5 \hat{k})] \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})-(\hat{\imath}-2 \hat{\jmath}+5 \hat{k}) \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})=0 \\ &\Rightarrow \vec{r}-(3 \hat{\imath}+\hat{\jmath}-\hat{k})-[(1)(3)+(-2)(1)+(5)(-1)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})-[3-2-5]=0 \\ &\Rightarrow \vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0 \end{aligned}$
$\Rightarrow \vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0$ is the vector equation of a required plane
Then, the above vector equation of the plane becomes
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(3)+(y)(1)+(z)(-1)=-4 \end{aligned}$
$\Rightarrow 3 x+y-z=-4$ is the cartesian form of equation of the required plane

The Plane exercise 28.3 question 12

Answer:
The answer of given question is $2x+2y+2z=18$
Hint:
$\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2}$
Given:
The given plane bisects the line segment joining points $A(1,2,3)$ and $B(3,4,5)$ and is at right angle to it..
Solution:
As the given plane bisects the line segment
Therefore,
$\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2}$
$\begin{aligned} &\Rightarrow \vec{a}=\frac{(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})}{2} \\ &\Rightarrow \vec{a}=\frac{4 \hat{\imath}+6 \hat{\jmath}+8 \hat{k}}{2} \\ &\Rightarrow \vec{a}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \ldots(i) \end{aligned}$
And it is also given the plane B normal to the line joining the points $A(1,2,3)$ and $B(3,4,5)$
Then $\vec{n}=\overrightarrow{A B}$
? $\vec{n}$ = position vector of $\vec{B}$ $-$ Position vector of $\vec{A}$
$\begin{aligned} &\vec{n}=(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &\vec{n}=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \ldots(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})] \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}) \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-[(2)(2)+(3)(2)+(4)(2)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-[4+6+8]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-18=0 \\ &\Rightarrow 2 x+2 y+2 z=18 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 13(i)

Answer:
The normal to the given pairs of planes are perpendicular to each other.
Hint:
$\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=0$
Given:
$\begin{aligned} &x-y+z-2=0 \\ &3 x+2 y-z+4=0 \end{aligned}$
Solution:
The vector equation of the planes $x-y+z-2=0$ can be written as
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}-\hat{\jmath}+\hat{k})=2 \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=2 \end{aligned}$
The normal to this plane is $\overrightarrow{n_{1}}=\hat{\imath}-\hat{\jmath}+\hat{k} \ldots(\mathrm{i})$
The vector equation of the plane $3 x+2 y-z+4=0$ can be written as
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(3 \hat{\imath}+2 \hat{\jmath}-\hat{k})=-4 \\ &\Rightarrow \vec{r} \cdot(3 \hat{\imath}+2 \hat{\jmath}-\hat{k})=-4 \end{aligned}$
The normal to this plane is
$\overrightarrow{n_{2}}=3 \hat{\imath}+2 \hat{\jmath}-\hat{k} \ldots(ii)$
Now,
$\begin{aligned} &\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(\hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(3 \hat{\imath}+2 \hat{\jmath}-\hat{k}) \\ &\Rightarrow \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(1)(3)+(-1)(2)+(1)(-1)=0 \end{aligned}$
Hence $\overrightarrow{n_{1}}$ is perpendicular to $\overrightarrow{n_{2}}$
Therefore, the normal to the given pair of planes are perpendicular to each other.

The Plane exercise 28.3 question 13(ii)

Answer:
The normal to the given pairs of planes are perpendicular to each other.
Hint:
$\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=0$
Given:
$\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})=5 \text { and } \vec{r} \cdot(2 \hat{\imath}-2 \hat{\jmath}-2 \hat{k})=5$
Solution:
The equation of the first plane is
$\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})=5$
The normal to this plane is
$\overrightarrow{n_{1}}=2 \hat{\imath}-\hat{\jmath}+3 \hat{k} \ldots(i)$
The equation of the second plane is
$\vec{r} \cdot(2 \hat{\imath}-2 \hat{\jmath}-2 \hat{k})=5$
The normal to this plane is
$\overrightarrow{n_{2}}=2 \hat{\imath}-2 \hat{\jmath}-2 \hat{k} \ldots(ii)$
Now,
$\begin{aligned} &\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}) \cdot(2 \hat{\imath}-2 \hat{\jmath}-2 \hat{k}) \\ &\Rightarrow \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(2)(2)+(-1)(-2)+(3)(-2)=0 \end{aligned}$
Hence $\overrightarrow{n_{1}}$ is perpendicular to $\overrightarrow{n_{2}}$
Therefore, the normal to the given pair of planes are perpendicular to each other.

The Plane exercise 28.3 question 14

Answer:
The answer of given question is the normal vector to the plane $2x+2y+2z=3$ is equally inclined with the co-ordinate axes.
Hint:
Let α,β,γ be the angle that normal $\vec{n}$ makes with the co-ordinate axes
Given:
$2x+2y+2z=3$
Solution:
The vector equation of the plane $2x+2y+2z=3$ can be written as
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=3 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=3 \end{aligned}$
The normal to this plane is
$\vec{n}=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \ldots(i)$
Direction ratio of $\vec{n}=2,2,2$
Direction cosine of
$\begin{aligned} &\vec{n}=\frac{2}{|\vec{n}|}, \frac{2}{|\vec{n}|}, \frac{2}{|\vec{n}|} \\ &\Rightarrow|\vec{n}|=\sqrt{(2)^{2}+(2)^{2}+(-2)^{2}} \\ &\Rightarrow|\vec{n}|=\sqrt{4+4+4} \\ &\Rightarrow|\vec{n}|=2 \sqrt{3} \end{aligned}$
Direction cosine of
$|\vec{n}|=\frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
So,
$l=\frac{1}{\sqrt{3}}, m=\frac{1}{\sqrt{3}}, n=\frac{1}{\sqrt{3}}$
Let α,β,γ be the angle that normal $\vec{n}$ makes with the co-ordinate axes respectively.
$\begin{aligned} &l=\cos \alpha=\frac{1}{\sqrt{3}} \\ &\Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i i) \\ &\Rightarrow \beta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i i i) \\ &\Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i v) \end{aligned}$
Hence α = β = γ
So the normal vector to the plane $2x+2y+2z=3$ is equally inclined with the co-ordinate axes.

12 chapter The Plane exercise 28.3 question 15

Answer:
The answer of given question is $24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k}$
Hint:
$\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})$
Given:
$12x-3y+4z=1$
Solution:
The vector equation of the plane $12x-3y+4z=1$ can be written as
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \\ &\Rightarrow \vec{r} \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \end{aligned}$
The normal to this plane is
$\begin{aligned} &\vec{n}=12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} \ldots(i) \\ &|\vec{n}|=\sqrt{(12)^{2}+(-3)^{2}+(4)^{2}} \\ &\Rightarrow|\vec{n}|=\sqrt{144+9+16} \\ &\Rightarrow|\vec{n}|=\sqrt{169} \\ &=13 \end{aligned}$
The unit vector becomes
$\begin{aligned} &\vec{n}=\frac{12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}}{13} \\ &=\frac{12}{13} \hat{\imath}-\frac{3}{13} \hat{\jmath}+\frac{4}{13} \hat{k} \end{aligned}$
Now a vector normal to the plane with the magnitude 26 will be
$\begin{aligned} &=26 \vec{n} \\ &\Rightarrow 26\left(\frac{12}{13} \hat{\imath}-\frac{3}{13} \hat{\jmath}+\frac{4}{13} \hat{k}\right) \\ &\Rightarrow 24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k} \end{aligned}$
A vector of magnitude 26 units normal to the plane is $24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k}$

The Plane exercise 28.3 question 16

Answer:
The answer of given question is $-14x+6y+2z=178$ or $-7x+3y+z=89$
Hint:
By using formula $\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
$A(4,-1,2)$ and $B(-10, 5,4).$
Solution:
It means that the plane is passing through $B(-10, 5,4).$
Therefore, the position vector of this point is,
$\Rightarrow \vec{a}=-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k} \ldots(i)$
And also given the line segment joining the points $A(4,-1,2)$ and $B(-10, 5,4).$ and is at right angle to it.
Then $\vec{n}=\overrightarrow{A B}$
? $\vec{n}$ =position vector of $\vec{B}$ $-$ Position vector of $\vec{A}$
$\begin{aligned} &\vec{n}=(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k})-(4 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &\vec{n}=-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k} \ldots(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k})] \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}) \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-[(-10)(-14)+(5)(6)+(4)(2)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-[140+30+8]=0 \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-178=0 \end{aligned}$
$\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=178$ is the vector equation of a required plane.
$\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})$
Then the above vector equation of a plane becomes,
$\begin{aligned} &\Rightarrow(x)(-14)+(y)(6)+(z)(2)=178 \\ &\Rightarrow-14 x+6 y+2 z=178 \text { or }-7 x+3 y+z=89 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 17

Answer:
$\vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28$ is the vector equation of a required plane &
$4 x-7 y+3 z=28$ is the cartesian form of equation of the required plane.
Hint:
By using formula $\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
$(-1,2,3)\; and \; (3,-5,6).$
Solution:
It means that the plane is passing through the midpoint of the line AB.
$\begin{aligned} &\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2} \\ &\Rightarrow \vec{a}=\frac{(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+(3 \hat{\imath}-5 \hat{\jmath}+6 \hat{k})}{2} \\ &\Rightarrow \vec{a}=\frac{2 \hat{\imath}-3 \hat{\jmath}+9 \hat{k}}{2} \\ &\Rightarrow \vec{a}=\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k} \ldots(i) \end{aligned}$
And also given the plane B normal to the line joining the points $A(-1,2,3) \; and \; B(3,-5,6)$
Then $\vec{n}=\overrightarrow{A B}$
? $\vec{n}$ = position vector of $\vec{B}$ $-$ Position vector of $\vec{A}$
$\begin{aligned} &\vec{n}=(3 \hat{\imath}-5 \hat{\jmath}+6 \hat{k})-(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &\vec{n}=4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k} \ldots(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get $\begin{aligned} &{\left[\vec{r}-\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k}\right)\right] \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k}\right) \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[(1)(4)+\left(-\frac{3}{2}\right)(-7)+\left(\frac{9}{2}\right)(3)\right]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
$\begin{aligned} &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[4+\frac{21}{2}+\frac{27}{2}\right]=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[\frac{8+21+27}{2}\right]=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-28=0 \end{aligned}$
$\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28$ is the vector equation of a required plane.
$\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})$
Then the above vector equation of a plane becomes,
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28$
Now multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(4)+(y)(-7)+(z)(3)=28 \\ &\Rightarrow 4 x-7 y+3 z=28 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 18

Answer:
The answer of given question is $2x+3y-z=20$
Hint:
By using formula $\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
The plane is passing through $P(5,2,-4)$ and perpendicular to the line having $2,3,-1$ as the direction ratio.
Solution:
Let, the position vector of this point P be
$\Rightarrow \vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \ldots(i)$
And also given the plane is normal having $2,3,-1$
As the direction ratio.
$\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots(ii)$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})] \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[(2)(5)+(2)(3)+(-4)(-1)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[10+6+4]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-20=0 \end{aligned}$
$\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=20$ is the vector equation of a required plane.
$\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})$
Then the above vector equation of a plane becomes,
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=20$
Now multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(2)+(y)(3)+(z)(-1)=20 \\ &\Rightarrow 2 x+3 y-z=20 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 19

Answer:
The answer of given question is $x+2y-3z=14$
Hint:
By using formula $\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
The plane is passing through $P(1,2,-3)$ and perpendicular to OP
Solution:
The required plane is passing through $P(1,2,-3)$ and perpendicular to OP. let the position vector of this point P be
$\Rightarrow \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \ldots(i)$
And it is also given the planes is normal to the line joining the points $O(0,0,0)$ and position vector of point $P(1,2,-3).$
Then
$\vec{n}=\overrightarrow{O P}$
? $\vec{n}$ = position vector of $\vec{P}$ $-$ Position vector of $\vec{A}$
$\begin{aligned} &\vec{n}=(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\vec{n}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \ldots .(i i) \end{aligned}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})] \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=0 \\ &\Rightarrow \vec{r}-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-[(1)(1)+(2)(2)+(-3)(-3)]=0 \end{aligned}$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-[1+4+9]=0 \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-14=0 \end{aligned}$ is the vector equation of the required plane
$\operatorname{Let} \vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})$
Then, the above vector equation of the plane becomes
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=14$
Now multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(1)+(y)(2)+(z)(-3)=14 \\ &\Rightarrow x+2 y-3 z=14 \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 20

Answer:
The answer of given question is $a x+b y+c z=a^{2}+b^{2}+c^{2}$
Hint:
$\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$
Given:
The required plane is passing through A(a,b,c) and perpendicular to OA.
Solution:
Let the position vector of this point A be
$\vec{a}=a \hat{\imath}+b \hat{\jmath}+c \hat{k} \ldots(i)$
And it is also given the planes is normal to the line joining the points O(0,0,0) and position vector of point A(a,b,c)
Then
$\vec{n}=\overrightarrow{O A}$
$\Rightarrow \vec{n}$ = position vector of $\vec{A}$ - Position vector of $\vec{O}$
$\begin{aligned} &\Rightarrow \vec{n}=(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\Rightarrow \vec{n}=a \hat{\imath}+b \hat{\jmath}+c \hat{k} \ldots .(i i) \end{aligned}$
The direction ratio of OA are proportional to a,b,c.
Hence, the direction cosines are
$\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
We know that
$(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Substituting the values from equation (i) and equation (ii) in the above equation we get
$\begin{aligned} &{[\vec{r}-(a \hat{\imath}+b \hat{\jmath}+c \hat{k})] \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=0 \end{aligned}$
$\Rightarrow \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-\left[a^{2}+b^{2}+c^{2}\right]=0$
By multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=a^{2}+b^{2}+c^{2} \\ &\text { Let } \vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \end{aligned}$
Then, the above vector equation of the plane becomes
$(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=a^{2}+b^{2}+c^{2}$
Now multiplying the two vectors using the formula
$\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(a)+(y)(b)+(z)(c)=a^{2}+b^{2}+c^{2} \\ &\Rightarrow a x+b y+c z=a^{2}+b^{2}+c^{2} \end{aligned}$
This is the cartesian form of equation of the required plane.

The Plane exercise 28.3 question 21

Answer:
The answer of given question is $\vec{r} \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12$
Hint:
Let the equation be $Ax+By+Cz+D=0$
Given:
Plane with intercepts 3,-4 and 2 on x, y and z axes
Solution:
Let the equation be $Ax+By+Cz+D=0$ ….(i)
Let the plane meets the x,y,z axes (3,0,0), (0,-4,0) and (0,0,2) respectively.
Putting (0,0,2), we get
$\begin{aligned} &A(0)+B(0)+C(2)+D=0 \\ &\Rightarrow C=-\frac{D}{2} \end{aligned}$
Putting (0,-4,0), we get
$\begin{aligned} &A(0)+B(-4)+C(0)+D=0 \\ &\Rightarrow B=\frac{D}{4} \end{aligned}$
And putting (3,0,0) we get
$\begin{aligned} &A(3)+B(0)+C(0)+D=0 \\ &\Rightarrow A=-\frac{D}{3} \end{aligned}$
Substituting the values of A, B, C in equation (i) we get by putting B(0,-4,0), we get
$\begin{aligned} &\left(-\frac{D}{3}\right) x+\frac{D}{4} y+\left(-\frac{D}{2}\right) z+D=0 \\ &\Rightarrow \frac{-4 D x+3 D y-6 D z+12 D}{2}=0 \\ &\Rightarrow(-D)(4 x-3 y+6 z-12)=0 \\ &\Rightarrow 4 x-3 y+6 z=12 \end{aligned}$
This is the cartesian form of equation of the required plane.
Now vector equation form of the plane $4x-3y+6z=12$
$\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12 \end{aligned}$

RD Sharma class 12 solutions The Plane 28.3 is one NCERT solution that is a must-have for all class 12 students who have maths in their board exams. This chapter covers concepts like Direction cosines and ratios, vector equation of a plane, a cartesian form of the equation, etc. Exercise 28.3 has 25 questions that are based on fundamentals and are relatively more straightforward. The RD Sharma class 12th exercise 28.3 has many questions which can be practiced by students to perfect their maths skills.

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The ideas that are covered under the RD Sharma class 12 solutions The Plane ex 28.3 books are Direction cosines and proportions, condition of Plane, Coplanarity of 2 lines, Equation of line under planes condition, the distance between slant lines, and points between two lines.

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The Plane Excercise: 28.3

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