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RD Sharma Class 12 Exercise FBQ The Plane Solutions Maths - Download PDF Free Online
RD Sharma books are the best books with regards to planning for board exams. However, board exams have consistently been viewed as a fabrication for some students. RD Sharma solutions In any case, when they allude to this book alongside Rd Sharma Class 12th Exercise FBQ solutions, they think that it is simpler to back out with a great deal of pressure and comprehend the idea in a vastly improved manner. This specific exercise has 20 inquiries from the general chapter. Rd Sharma Class 12th Exercise FBQ incorporates Equation of the plane containing two lines, Equation of the plane in scalar item structure, Reflection of point, Vector and Cartesian structure, condition of the plane in standard form, Shortest distance between two lines, and distance between the plane.
This Story also Contains
- RD Sharma Class 12 Solutions Chapter28 FBQ The Plane - Other Exercise
- The Plane Exercise: FBQ
- RD Sharma Chapter-wise Solutions
RD Sharma Class 12 Solutions Chapter28 FBQ The Plane - Other Exercise
The Plane Exercise: FBQ
The Plane exercise fill in the blank question 1
Answer:10
Hint:
Use vector dot product
Given:
x + 2y - 2z = d, d>0
Solution:
Perpendicular distance of the plane ax +by + cz + d = 0
$(x,y,z) \text { is } d = \left | \frac{ax_{1}+by_{1}+cz_{1}+d}{a^{2}+b^{2}+c^{2}} \right |$
Equation of plane parallel to x + 2y - 2z = d is
x + 2y -2z - d = 0 …………………………(1)
Perpendicular distance from (1,-2,1) is S
$\text { S } = \left | \frac{1-4-2+d}{\sqrt{1+4+4}} \right |\Rightarrow d=10$
The Plane exercise fill in the blank question 2
Answer:
10
Hint:
Two perpendicular plane equals 0
Given:
3x - 6y - 2z = 7 and 2x + y - λz = 5
Solution:
The equations of planes are
3x - 6y - 2z -7 =0
and 2x + y - kz - 5 = 0
Since two planes are perpendicular to each other
∴(3)(2) + (-6)(1) + (-2) (-k) = 0 ? a1a2+b1b2+c1c2=0
∴6 - 6 + 2k = 0 2k = 0
k = 0
10
Hint:
Two perpendicular plane equals 0
Given:
3x - 6y - 2z = 7 and 2x + y - λz = 5
Solution:
The equations of planes are
3x - 6y - 2z -7 =0
and 2x + y - kz - 5 = 0
Since two planes are perpendicular to each other
∴(3)(2) + (-6)(1) + (-2) (-k) = 0 ? a1a2+b1b2+c1c2=0
∴6 - 6 + 2k = 0 2k = 0
k = 0
The Plane exercise fill in the blank question 3
Answer:
2X + 3Y - Y2 + 4 = 0
Hint:
Parallel plane equation also equals to 0
Given:
(1,2,3) and parallel plane 2x + 3y - 4 = 0
Solution:
Equation of parallel to2x + 3y - y2 = 0 is given by
2x + 3y - y2 + λ = 0
Also It passes through (1,2,3)
? 2(1) + 3(2) - 4(3) + λ = 0 ? λ = 4
∴ required plane is 2x + 3y - y2 + 4 = 0
2X + 3Y - Y2 + 4 = 0
Hint:
Parallel plane equation also equals to 0
Given:
(1,2,3) and parallel plane 2x + 3y - 4 = 0
Solution:
Equation of parallel to2x + 3y - y2 = 0 is given by
2x + 3y - y2 + λ = 0
Also It passes through (1,2,3)
? 2(1) + 3(2) - 4(3) + λ = 0 ? λ = 4
∴ required plane is 2x + 3y - y2 + 4 = 0
The Plane exercise fill in the blank question 4
Answer:
$\frac{X-1}{2}=\frac{Y-2}{3}=\frac{Z-3}{6}$
Hint:
Use bracket method
Given:
(1,2,3) and normal to the plane 2x - 3y + 6z = 11
Solution:
Let the equation be
$\frac{X-1}{2}=\frac{Y-2}{3}=\frac{Z-3}{6} = 0$
$\sqrt{(1)^{2}+\left ( \frac{3}{2} \right )+\left ( \frac{6}{3} \right )}$
? we will get normal equation
? 2x - 3y + 6z = 11
$\frac{X-1}{2}=\frac{Y-2}{3}=\frac{Z-3}{6}$
Hint:
Use bracket method
Given:
(1,2,3) and normal to the plane 2x - 3y + 6z = 11
Solution:
Let the equation be
$\frac{X-1}{2}=\frac{Y-2}{3}=\frac{Z-3}{6} = 0$
$\sqrt{(1)^{2}+\left ( \frac{3}{2} \right )+\left ( \frac{6}{3} \right )}$
? we will get normal equation
? 2x - 3y + 6z = 11
The Plane exercise fill in the blank question 5
Answer:
$k=\frac{1}{2}$
Hint:
Use vector 3 dimension
Given:
2x + 3y - 4k = 24 on the coordinate is 8
Solution:
$\therefore$ sum of intercepts
=2 + 3 - 4k = 24
$k=\frac{1}{2}$
Hint:
Use vector 3 dimension
Given:
2x + 3y - 4k = 24 on the coordinate is 8
Solution:
$\therefore$ sum of intercepts
=2 + 3 - 4k = 24
The Plane exercise fill in the blank question 6
Answer:
3, -9
Hint:
Solve using fraction method.
Given:
(1,1,1) is half distance fromx + 2y + z + k = 0
Solution:
$\begin{aligned} &\sqrt{3}=\frac{1}{2}\left [ \frac{3+k}{\sqrt{3}} \right ]\\ &3=\pm \frac{1}{2}(3+k)\\ &3+k=\pm 6\\ &3+k=6 \qquad 3+k=6\\ &k=3 \qquad \qquad k=-9 \end{aligned}$
3, -9
Hint:
Solve using fraction method.
Given:
(1,1,1) is half distance fromx + 2y + z + k = 0
Solution:
$\begin{aligned} &\sqrt{3}=\frac{1}{2}\left [ \frac{3+k}{\sqrt{3}} \right ]\\ &3=\pm \frac{1}{2}(3+k)\\ &3+k=\pm 6\\ &3+k=6 \qquad 3+k=6\\ &k=3 \qquad \qquad k=-9 \end{aligned}$
The Plane exercise fill in the blank question 7
Answer:
(a,b,c)
Hint:
Let the points a,b,c and try to get the equation
Given:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3$
Solution:
Let the point be (a,b,c)
The plane meets the coordinates are at A,B and C such that centroid of the triangle ABC is (a,b,c)
So, the plane cuts x- axis at (3a,0,0) , so, x- intercept = 3a
The plane cuts z- axis at (0,0,3c)?z- intercept =3c
The plane cuts y-axis at (0,3b,0)?y- intercept =3b
? all one getting the equation
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3$
∴ the point will be (a,b,c)
The Plane exercise fill in the blank question 8
Answer:
ax + by + cz = a2 + b2 + c2
Hint:
Firs find the co-ordinates
Given:
p(a,b,c) perpendicular to op
Solution:
The coordinates of the points o and p ,are (0,0,0) and (a,b,c) respectively. Therefore the direction rahos of op are (a-0)=a,(b-0) = b, and (c - 0) = c
It is known that the equation of the plane passing through the point (x,y,z) is
a(x - x1) + b(y - y1)+ c(z - z1) = 0
∴ the required equation will be
a(x - a) + b(y - b) + c(z - c) = 0
ax - a2 + by - b2 + cz - c2 = 0
ax + by + cz = a2 + b2 + c2
ax + by + cz = a2 + b2 + c2
Hint:
Firs find the co-ordinates
Given:
p(a,b,c) perpendicular to op
Solution:
The coordinates of the points o and p ,are (0,0,0) and (a,b,c) respectively. Therefore the direction rahos of op are (a-0)=a,(b-0) = b, and (c - 0) = c
It is known that the equation of the plane passing through the point (x,y,z) is
a(x - x1) + b(y - y1)+ c(z - z1) = 0
∴ the required equation will be
a(x - a) + b(y - b) + c(z - c) = 0
ax - a2 + by - b2 + cz - c2 = 0
ax + by + cz = a2 + b2 + c2
The Plane exercise fill in the blank question 9
Answer:
(2,2,2)
Hint:
Let its intercept be λ
Given:
Sum of the reciprocals of its intercepts on the coordinates axis is 1/2
Solution:
Let the equation of the variable plane be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\lambda \dots \dots \dots \dots(1)$
The intercepts on the coordinate axes are a,b,c
Let the sum of reciprocals of intercept in constant =λ
Therefore,
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\lambda \Rightarrow \frac{(\frac{1}{\lambda })}{a}+\frac{(\frac{1}{\lambda })}{b}+\frac{(\frac{1}{\lambda })}{c}=\lambda$
$\left ( \frac{1 }{\lambda },\frac{1 }{\lambda },\frac{1 }{\lambda } \right )$
lies on the plane (1)
Hence, the variable plane (1) always passes through
the fixed point
$\left ( \frac{1 }{(\frac{1}{2})},\frac{1 }{(\frac{1}{2})},\frac{1 }{(\frac{1}{2})}\Rightarrow 2,2,2 \right )$
(2,2,2)
Hint:
Let its intercept be λ
Given:
Sum of the reciprocals of its intercepts on the coordinates axis is 1/2
Solution:
Let the equation of the variable plane be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\lambda \dots \dots \dots \dots(1)$
The intercepts on the coordinate axes are a,b,c
Let the sum of reciprocals of intercept in constant =λ
Therefore,
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\lambda \Rightarrow \frac{(\frac{1}{\lambda })}{a}+\frac{(\frac{1}{\lambda })}{b}+\frac{(\frac{1}{\lambda })}{c}=\lambda$
$\left ( \frac{1 }{\lambda },\frac{1 }{\lambda },\frac{1 }{\lambda } \right )$
lies on the plane (1)
Hence, the variable plane (1) always passes through
the fixed point
$\left ( \frac{1 }{(\frac{1}{2})},\frac{1 }{(\frac{1}{2})},\frac{1 }{(\frac{1}{2})}\Rightarrow 2,2,2 \right )$
The Plane exercise fill in the blank question 10
Answer:
1
Hint:
Use formulae of circumantce after finding c0-ordinates
Given:
x + αy +2 = 5
Solution:
Given equation of plane
$x + \alpha y +2=5\Rightarrow \frac{x}{5}+\frac{\alpha y}{5}+\frac{z}{5}=1$
Comparing above equation with equation of plane in intercept from
$\frac{x}{x_{1}}+\frac{y}{y_{1}}+\frac{z}{z_{1}}=1$
$x_1=5,y_1=5,z_1=5$
So coordinates of points A,B,C?A(5,0,0),B(0,5,0),C(0,0,5)
Let circumentre D(x,y,z)
$\therefore AD=\sqrt{(x-5)^{2}+y^{2}+z^{2}}, BD=\sqrt{x^{2}+(y-5)^{2}+z^{2}}, CD=\sqrt{x^{2}+y^{2}+(z-5)^{2}}$
AD=BD=CD ? taking AD=BD
$\Rightarrow \sqrt{(x-5)^{2}+y^{2}+z^{2}}=\sqrt{x^{2}+(y-5)^{2}+z^{2}}$
On squaring both sides
(x - 5)2 + y2 + z2 = x2 + (y - 5)2 + z2 ? 1 = 1
∴ from equation (1) ,x,y can be x=1, y=1
∴ α=1 (also)(by solving)
1
Hint:
Use formulae of circumantce after finding c0-ordinates
Given:
x + αy +2 = 5
Solution:
Given equation of plane
$x + \alpha y +2=5\Rightarrow \frac{x}{5}+\frac{\alpha y}{5}+\frac{z}{5}=1$
Comparing above equation with equation of plane in intercept from
$\frac{x}{x_{1}}+\frac{y}{y_{1}}+\frac{z}{z_{1}}=1$
$x_1=5,y_1=5,z_1=5$
So coordinates of points A,B,C?A(5,0,0),B(0,5,0),C(0,0,5)
Let circumentre D(x,y,z)
$\therefore AD=\sqrt{(x-5)^{2}+y^{2}+z^{2}}, BD=\sqrt{x^{2}+(y-5)^{2}+z^{2}}, CD=\sqrt{x^{2}+y^{2}+(z-5)^{2}}$
AD=BD=CD ? taking AD=BD
$\Rightarrow \sqrt{(x-5)^{2}+y^{2}+z^{2}}=\sqrt{x^{2}+(y-5)^{2}+z^{2}}$
On squaring both sides
(x - 5)2 + y2 + z2 = x2 + (y - 5)2 + z2 ? 1 = 1
∴ from equation (1) ,x,y can be x=1, y=1
∴ α=1 (also)(by solving)
The Plane exercise fill in the blank question 11
Answer:
-4
Hint:
Use vector dot product
Given:
$\frac{2x-1}{2}=\frac{2-y}{2}=\frac{2+z}{k}$
is parallel to the plane 2x - y + z = 3
Solution:
$\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\\ &\Rightarrow \frac{x-\frac{1}{2}}{1}=\frac{y-2}{-2}=\frac{z+1}{9}\\ &\vec{x}=2\hat{i}-\hat{j}+\hat{k}, \quad \vec{b}=\hat{i}-2\hat{j}+\hat{k}\\ &\vec{x}.\vec{b}=0\Rightarrow 2+2+k=0\Rightarrow k=-4 \end{aligned}$
-4
Hint:
Use vector dot product
Given:
$\frac{2x-1}{2}=\frac{2-y}{2}=\frac{2+z}{k}$
is parallel to the plane 2x - y + z = 3
Solution:
$\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\\ &\Rightarrow \frac{x-\frac{1}{2}}{1}=\frac{y-2}{-2}=\frac{z+1}{9}\\ &\vec{x}=2\hat{i}-\hat{j}+\hat{k}, \quad \vec{b}=\hat{i}-2\hat{j}+\hat{k}\\ &\vec{x}.\vec{b}=0\Rightarrow 2+2+k=0\Rightarrow k=-4 \end{aligned}$
The Plane exercise fill in the blank question 12
Answer:-6
Hint:
Use direction ratios
Given:
$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$
Solution:
The line is
$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$
The direction radius of the line are(3,-5,2)
As the line lies in the plane x + 3y - αz + 7 = 0
We have $3-15-2\alpha+7=0\Rightarrow -2\alpha-5=0$
$\\ \\ \alpha=-\frac{5}{2}$
The Plane exercise Fill in the blank question 13
Answer:
-9
Hint:
Use direction ratios
Given:
$\frac{x+1}{\lambda }=\frac{y-1}{1}=\frac{z+2}{-4}$
Solution:
Let the given line is
$\frac{x+1}{\lambda }=\frac{y-1}{1}=\frac{z+2}{-4}$
Direction ratios are λ,1,-4
And the given plane 2x+ 2y - 8z = -5
Its direction ratios are 2,2,-4
And the given line and plane are perpendicular, then
λ(1) + 1(1) + (-4)(-8) = 0
2λ + 2 + 16=0
λ = -9
-9
Hint:
Use direction ratios
Given:
$\frac{x+1}{\lambda }=\frac{y-1}{1}=\frac{z+2}{-4}$
Solution:
Let the given line is
$\frac{x+1}{\lambda }=\frac{y-1}{1}=\frac{z+2}{-4}$
Direction ratios are λ,1,-4
And the given plane 2x+ 2y - 8z = -5
Its direction ratios are 2,2,-4
And the given line and plane are perpendicular, then
λ(1) + 1(1) + (-4)(-8) = 0
2λ + 2 + 16=0
λ = -9
The Plane exercise Fill in the blank question 14
Answer:
6x + 4y + 3z = 12 is the required equation of plane
Hint:
Equation of the plane that cut the coordinate axes =1
Given:
(2,0,0) , (0,3,0) and (0,0,4)
Solution:
We know that, equation of the plane that cut the coordinate axes at
(2,0,0) , (0,3,0) and (0,0,4) is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Hence, the equation of plane passes through the points (2,0,0) , (0,3,0) and (0,0,4)
is
$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
= 6x + 4y + 3z = 12
6x + 4y + 3z = 12 is the required equation of plane
Hint:
Equation of the plane that cut the coordinate axes =1
Given:
(2,0,0) , (0,3,0) and (0,0,4)
Solution:
We know that, equation of the plane that cut the coordinate axes at
(2,0,0) , (0,3,0) and (0,0,4) is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Hence, the equation of plane passes through the points (2,0,0) , (0,3,0) and (0,0,4)
is
$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
= 6x + 4y + 3z = 12
The Plane exercise fill in the blank question 15
Answer:
x + y + z = 2 is the required Cartesian product
Hint:
Use vector product
Given:
$\vec{r}.\left ( \hat{i}+\hat{j}+\hat{k} \right )=2$
Solution:
We have,
$\begin{aligned} &\vec{r}.\left ( \hat{i}+\hat{j}+\hat{k} \right )=2\\ &\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\left ( \hat{i}+\hat{j}+2 \right )=2\\ &\Rightarrow x+y+z=2 \end{aligned}$
x + y + z = 2 is the required Cartesian product
Hint:
Use vector product
Given:
$\vec{r}.\left ( \hat{i}+\hat{j}+\hat{k} \right )=2$
Solution:
We have,
$\begin{aligned} &\vec{r}.\left ( \hat{i}+\hat{j}+\hat{k} \right )=2\\ &\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\left ( \hat{i}+\hat{j}+2 \right )=2\\ &\Rightarrow x+y+z=2 \end{aligned}$
The Plane exercise Fill in the blank question 16
Answer:
$-2,\frac{4}{3},\frac{-4}{5}$
are the intercepts made by the plane
Hint:
Convert the given equation into intercept form
Given:
2x - 3y + 5z + 4 = 0
Solution:
First let’s convert the given equation to intercept form i.e
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Where a,b,c are x,y nad z intercepts respectively
Given,
2x - 3y + 5z + 4 = 0
⇒ -2x + 3y - 5z = 4
Dividing by 4 both sides
$\frac{x}{-2}+\frac{y}{\frac{4}{3}}+\frac{z}{\frac{4}{-5}}=1$
On comparing we have intercepts as
$-2,\frac{4}{3},\frac{-4}{5}$
$-2,\frac{4}{3},\frac{-4}{5}$
are the intercepts made by the plane
Hint:
Convert the given equation into intercept form
Given:
2x - 3y + 5z + 4 = 0
Solution:
First let’s convert the given equation to intercept form i.e
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Where a,b,c are x,y nad z intercepts respectively
Given,
2x - 3y + 5z + 4 = 0
⇒ -2x + 3y - 5z = 4
Dividing by 4 both sides
$\frac{x}{-2}+\frac{y}{\frac{4}{3}}+\frac{z}{\frac{4}{-5}}=1$
On comparing we have intercepts as
$-2,\frac{4}{3},\frac{-4}{5}$
The Plane exercise Fill in the blank question 17
Answer:
$sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]$
is the angle between the line and the plane
Hint:
Find $\theta$ using vector dot product
Given:
$\vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right ) \text { and the plane }\vec{r}. \left ( 3\hat{i}-4\hat{j}-\hat{k} \right )+5=0$
Solution:
Line
$\vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right )$
Is parallel to the vector
$\vec{b}= 2\hat{i}-\hat{j}+\hat{k}$
Normal to the plane is
$\vec{n}= 3\hat{i}-4\hat{j}-\hat{k}$
Let $\theta$ is the angle between line and plane
Then,
$sin\theta =\frac{\left | \vec{b}.\vec{n} \right |}{\left | \vec{b} \right |\left | \vec{n} \right |}\Rightarrow \frac{\left | \left ( 2\hat{i}-\hat{j}+\hat{k} \right ).\left ( 3\hat{i}-4\hat{j}+\hat{k} \right ) \right |}{\sqrt{6}.\sqrt{26}}$
$=\frac{\left | 6+4-1 \right |}{\sqrt{156}}=\frac{9}{\sqrt{156}}$
$\theta = sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]$
$sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]$
is the angle between the line and the plane
Hint:
Find $\theta$ using vector dot product
Given:
$\vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right ) \text { and the plane }\vec{r}. \left ( 3\hat{i}-4\hat{j}-\hat{k} \right )+5=0$
Solution:
Line
$\vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right )$
Is parallel to the vector
$\vec{b}= 2\hat{i}-\hat{j}+\hat{k}$
Normal to the plane is
$\vec{n}= 3\hat{i}-4\hat{j}-\hat{k}$
Let $\theta$ is the angle between line and plane
Then,
$sin\theta =\frac{\left | \vec{b}.\vec{n} \right |}{\left | \vec{b} \right |\left | \vec{n} \right |}\Rightarrow \frac{\left | \left ( 2\hat{i}-\hat{j}+\hat{k} \right ).\left ( 3\hat{i}-4\hat{j}+\hat{k} \right ) \right |}{\sqrt{6}.\sqrt{26}}$
$=\frac{\left | 6+4-1 \right |}{\sqrt{156}}=\frac{9}{\sqrt{156}}$
$\theta = sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]$
The Plane exercise Fill in the blank question 18
Answer:
$\vec{r}.(5\hat{i}-3\hat{j}-2\hat{k})=38$
is the required equation of plane.
Hint:
Use figure to solve the equation
Given:
(5,-3,-2)
Solution:
![]()
From the figure normal to the plane is
$\vec{n}=\vec{OP}=(5\hat{i}-3\hat{j}-2\hat{k})$
Plane passing through the point P(5,-3,-2)
Equation of plane is
$\begin{aligned} &5(x-5)-3(y+3)-2(z+2)=0\\ &5x-3y-2z=38 \end{aligned}$
In vector form
$\vec{r}.(5\hat{i}-3\hat{j}-2\hat{k})=38$
$\vec{r}.(5\hat{i}-3\hat{j}-2\hat{k})=38$
is the required equation of plane.
Hint:
Use figure to solve the equation
Given:
(5,-3,-2)
Solution:
From the figure normal to the plane is
$\vec{n}=\vec{OP}=(5\hat{i}-3\hat{j}-2\hat{k})$
Plane passing through the point P(5,-3,-2)
Equation of plane is
$\begin{aligned} &5(x-5)-3(y+3)-2(z+2)=0\\ &5x-3y-2z=38 \end{aligned}$
In vector form
$\vec{r}.(5\hat{i}-3\hat{j}-2\hat{k})=38$
The Plane exercise Fill in the blank question 19
Answer:
3 units is the required distance.
Hint:
use vector dot product to find the distance
Given:
$2x+y-2z-6=0 \text { and } 4x+2y-4z=0$
Solution:
Let
$\begin{aligned} &P_{1}\rightarrow 2x+y-2z-6=0\\ &ax+by+cz+d_{1}=0\\ &P_{2}\rightarrow 4x+2y-4z=0\\ &ax+by+cz+d_{2}=0\\ \end{aligned}$
$\begin{aligned} &\because P_{1}\left | \right |P_{2} \end{aligned}$
$\begin{aligned} &\text { Distance between } P_{1} \text { and } P_{2}=\left | \frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |\\ &=\left | \frac{0-6}{\sqrt{2^{2}+1^{2}+(-2)^{2}}} \right |=\left | \frac{-6}{4+1+4} \right |\\ &=3\: units \end{aligned}$
3 units is the required distance.
Hint:
use vector dot product to find the distance
Given:
$2x+y-2z-6=0 \text { and } 4x+2y-4z=0$
Solution:
Let
$\begin{aligned} &P_{1}\rightarrow 2x+y-2z-6=0\\ &ax+by+cz+d_{1}=0\\ &P_{2}\rightarrow 4x+2y-4z=0\\ &ax+by+cz+d_{2}=0\\ \end{aligned}$
$\begin{aligned} &\because P_{1}\left | \right |P_{2} \end{aligned}$
$\begin{aligned} &\text { Distance between } P_{1} \text { and } P_{2}=\left | \frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |\\ &=\left | \frac{0-6}{\sqrt{2^{2}+1^{2}+(-2)^{2}}} \right |=\left | \frac{-6}{4+1+4} \right |\\ &=3\: units \end{aligned}$
The Plane exercise Fill in the blank question 20
Answer:
x - 3z = 10 is the required Cartesian equation
Hint:
use direction cosine formula.
Given:
P(1,0,-3)
Solution:
Assume direction cosine of the normal (a,b,c)
Equation of plane passing through (x,y,z)
$\begin{aligned} &\Rightarrow a(x-x_{1})+b(y-y_1)+c(z-z_1)=0\\ \end{aligned}$
Point P(1,0,-3)
$\begin{aligned} &\Rightarrow a(x-1)+b(y-0)+c(z-(-3))=0\\ \end{aligned}$
Direction cosine of normal
(a,b,c) = (1 - 0, 0 - 0, -3 - 0)
(a,b,c) = (1,0,-3)
Equation of the plane
[(x - 1)+ 0 (y - 0) - 3 (2- (-3))]=0
x - 3z = 10
x - 3z = 10 is the required Cartesian equation
Hint:
use direction cosine formula.
Given:
P(1,0,-3)
Solution:
Assume direction cosine of the normal (a,b,c)
Equation of plane passing through (x,y,z)
$\begin{aligned} &\Rightarrow a(x-x_{1})+b(y-y_1)+c(z-z_1)=0\\ \end{aligned}$
Point P(1,0,-3)
$\begin{aligned} &\Rightarrow a(x-1)+b(y-0)+c(z-(-3))=0\\ \end{aligned}$
Direction cosine of normal
(a,b,c) = (1 - 0, 0 - 0, -3 - 0)
(a,b,c) = (1,0,-3)
Equation of the plane
[(x - 1)+ 0 (y - 0) - 3 (2- (-3))]=0
x - 3z = 10
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RD Sharma Chapter-wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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