RD Sharma Class 12 Ex 29.10 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.10 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 29.10 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.10
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 29.10 Solutions Chapter 29 The Plane

Question 1. Find the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0

Solution:

Let P(x1, y1, z1) be any point on plane 2x – y + 3z – 4 = 0.

⟹ 2x1 – y1 + 3z1 = 4 (equation-1)

Distance between (x1, y1, z1) and the plane

6x – 3y + 9z + 13 = 0:

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Now, substitute the values, we get

p = \left|\frac{(6)(x_1)+(-3)(y_1)+(9)(z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|

\left|\frac{3(2x_1-y_1+3z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|

 \left|\frac{3(4)+13}{3\sqrt{14}}\right|   [by using equation 1]

\frac{25}{3\sqrt{14}}

Therefore, the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0 is \frac{25}{3\sqrt{14}}  units.

Question 2. Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also find the distance between the two planes.

Solution:

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

It is given that,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

θ = -11

Thus, 

The equation of the plane is as follows:

2x – 3y + 5z – 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Now, after substituting the values, we will get

\left|\frac{(2)(3)+(-3)(4)+(5)(-1)+7}{\sqrt{2^2+(-3)^2+5^2}}\right|

\frac{4}{\sqrt{38}}

Therefore, the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, -1) is \frac{4}{\sqrt{38}}

Question 3. Find the equation of the plane mid-parallel to the planes 2x – 2y + z + 3 = 0 and 2x – 2y + z + 9 = 0

Solution:

Given:

Equation of planes: 

π1= 2x – 2y + z + 3 = 0

π2= 2x – 2y + z + 9 = 0

Let the equation of the plane mid–parallel to these planes be:

π3: 2x – 2y + z + θ = 0

Now,

Let P(x1, y1, z1) be any point on this plane,

⟹ 2(x1) – 2(y1) + (z1) + θ = 0 —(equation-1)

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Distance of P from π1:

p = \left|\frac{(2x_1-2y_2+z_1)+3}{\sqrt{2^2+(-2)^2+1^2}}\right|

=  \left|\frac{(-θ)+3}{3}\right|   (By using equation 1)

Similarly,

DIstance of q from π2:

q = \left|\frac{(2x_1-2y_2+z_1)+9}{\sqrt{2^2+(-2)^2+1^2}}\right|

\left|\frac{(-θ)+9}{3}\right|    (By using equation 1)

As π3 is mid-parallel is π1 and π2:

p = q 

So,

\left|\frac{(-θ)+3}{3}\right|=\left|\frac{-θ +9}{3}\right|

Now square on both sides, we get

\left(\frac{(-θ)+3}{3}\right)^2=\left(\frac{-θ+9}{3}\right)^2

(3 – θ)2 = (9 – θ)2

9 – 2×3×θ + θ2 = 81 – 2×9×θ + θ2

θ = 6

Now, substitute the value of θ = 6 in equation 2x – 2y + z + θ = 0, we get

Hence, the equation of the mid-parallel plane is 2x – 2y + z + 6 = 0

Question 4. Find the distance between the planes \vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0  and \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0

Solution:

Let \vec{a}  be the position vector of any pont P on the plane

\vec{r}(\hat{i}+2\hat{j}+3\hat{k})+7=0

So, 

\vec{a}(\hat{i}+2\hat{j}+3\hat{k})+7=0  —(equation 1)

As we know that, the distance of from  \vec{a}  the plane  \vec{r}.\vec{n}-d=0  is given by:

p = \left|\frac{\vec{a}.\vec{n}-d}{|n|}\right|

Length of perpendicular from P(\vec{a}) to plane \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0  is given by substituting the values of \vec{a}\  and\  \vec{n} , we get

p = \left|\frac{2\vec{a}.(2\hat{i}+4\hat{j}+6\hat{k})+7}{|2\hat{i}+4\hat{j}+6\hat{k}|}\right|

\left|\frac{2\vec{a}.(\hat{i}+2\hat{j}+3\hat{k})+7}{\sqrt{2^2+4^2+6^2}}\right|

\left|\frac{2(-7)+7}{\sqrt{56}}\right|

p = \frac{7}{\sqrt{56}}

Therefore, the distance between the planes

\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0  and  \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0  is \frac{7}{\sqrt{56}}

Question 14. Find the equation of the plane passing through the intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane.

Solution:

As we know that the equation of the plane passing through the line of intersection of two planes are

(a1x + b1y + c1z + d1) + λ(a2x + b2y + c2z + d2) = 0

So, the equation of the plane passes through the intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 is

 (1 + 2λ)x + (-2 + λ)y + (1 + λ)z – 1 – 8λ = 0          ….(1)

Also, given that this plane is parallel to the line whose direction ratios are proportional to 1, 2, 1.

⇒  (1 + 2λ)1 + (-2 + λ)2 + (1 + λ)1 = 0

⇒ 1 + 2λ – 4 + 2λ + 1 + λ = 0

⇒ 5λ – 2 = 0

⇒ λ = 2/5

Now put the value of λ in (1), we have:

9x – 8y + 7z – 21 = 0 is the required equation.

And the perpendicular distance of plane from (1, 1, 1) is

 = \frac{|9(1)-8(1)+7(1)-21|}{\sqrt{9^2+(-8)^2+7^2}}

\frac{13}{\sqrt{194}}   units.

Question 15. Show when the line \vec{r}=\vec{a}+λ\vec{b}   is parallel to the plane \vec{r}.\vec{n}=d    . Show that the line \vec{r}=\hat{i}+\hat{j}+λ(3\hat{i}-\hat{j}+2\hat{k})   is parallel to the plane \vec{r}.(2\hat{j}+\hat{k})=3   . Also, find the distance between the line and the plane.

Solution:

The plane passes through the point with position vector \vec{a}=\hat{i}+\hat{j}+0\hat{k}   and is parallel to the vector \vec{b}=3\hat{i}-\hat{j}+2\hat{k}.

Given equation of the plane is \vec{r}.(2\hat{j}+\hat{k})=3   or \vec{r}.\vec{n}=d.

So, the normal vector is \vec{n}=0\hat{i}+2\hat{j}+\hat{k}    and d = 3.

Now, \vec{b}.\vec{n}=(3\hat{i}-\hat{j}+2\hat{k}).(0\hat{i}+2\hat{j}+\hat{k})

= 0 – 2 + 2 = 0

So, \vec{b}   is perpendicular to \vec{n}

Hence, the given line is parallel to the given plane.

As we know that the distance between the line and parallel plane is the distance between any point on the line and the given plane. 

So, Distance(d) = \frac{|\vec{a}.\vec{n}-d|}{|\vec{n}|}

\frac{|(\hat{i}+\hat{j}+0\hat{k}).(0\hat{i}+2\hat{j}+\hat{k})-3|}{|0\hat{i}+2\hat{j}+\hat{k}|}

\frac{1}{\sqrt5}        units.

Question 16.  Show that the line \vec{r}=(-\hat{i}+2\hat{j}-\hat{k})+λ(2\hat{i}+\hat{j}+4\hat{k})    is parallel to the plane \vec{r}.(2\hat{i}+\hat{j}+4\hat{k})=3   . Also, find the distance between the two.

Solution:

The plane passes through the point with the position vector \vec{a}=\hat{i}+\hat{j}+0\hat{k}       and is parallel to the vector \vec{b}=3\hat{i}-\hat{j}+2\hat{k}.      [Tex]\vec{b}[/Tex]

Given plane is \vec{r}.(2\hat{j}+\hat{k})=3       or \vec{r}.\vec{n}=d.

So, the normal vector is \vec{n}=0\hat{i}+2\hat{j}+\hat{k}      and d = 3.

Now, \vec{b}.\vec{n}=(3\hat{i}-\hat{j}+2\hat{k}).(0\hat{i}+2\hat{j}+\hat{k})

= 0 – 2 + 2 = 0

So, it is perpendicular to \vec{n}

Hence, the given line is parallel to the plane.

As we know that the distance between a line and a parallel plane is the distance between any point on the line and the given plane.

so, Distance(d) = \frac{|\vec{a}.\vec{n}-d|}{|\vec{n}|}

\frac{|(\hat{i}+\hat{j}+0\hat{k}).(0\hat{i}+2\hat{j}+\hat{k})-3|}{|0\hat{i}+2\hat{j}+\hat{k}|}

= 1/√6 units.

Question 17. Find the equation of the plane through the intersection of the planes 3x – 4y + 5z = 10 and 2x + 2y – 3z = 4 and parallel to the line x = 2x = 3z.

Solution:

As we know that the equation of the plane passing through the line of intersection of two planes are

(a1x + b1y + c1z + d1) + λ(a2x + b2y + c2z + d2) = 0

So, the equation of the plane passes through the intersection of the 

planes 3x – 4y + 5z = 10 and 2x + 2y – 3z = 4 is 

 (3 + 2λ)x + (-4 + 2λ)y + (5 – 3λ)z – 10 – 4λ = 0          ….(1)

Given that the equation of line is x = 2x = 3z.

Now, dividing this equation by 6, we get,

\frac{x}{6}=\frac{y}{3}=\frac{z}{2}

So we get the direction ratios of this line are proportional to 6, 3, 2.

Now,  the normal to the plane is perpendicular to the line whose direction ratios are 6, 3, 2.

⇒  (3 + 2λ)6 + (-4 + 2λ)3 + (5 – 3λ)2 – 10 – 4λ = 0

⇒ λ = -4/3

Now put the value of λ in eq(1), we get,

x – 20y + 27z = 14 is the required equation.

Question 18. Find the vector and cartesian forms of the equation of the plane passing through the point (1, 2, -4) and parallel to the lines \vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})   and \vec{r}=(\hat{i}-3\hat{j}+5\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})  . Also, find the distance of the point from the plane obtained.

Solution:

Given that the equations of the lines are \vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})

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

As we know that the vector equation of a plane passing through a point \vec{a}  and parallel to \vec{b}      and \vec{c}      is 

(\vec{r}-\vec{a}).(\vec{b}×\vec{c})=0

So, \vec{a}=\hat{i}+2\hat{j}-4\hat{k} \vec{b}=2\hat{i}+3\hat{j}+6\hat{k} \vec{c}=\hat{i}+\hat{j}-\hat{k}

\vec{b}×\vec{c}=-9\hat{i}+8\hat{j}-\hat{k}

Now, the vector equation of the plane is (\vec{r}- \vec{a}).(\vec{b}\times \vec{c}) = 0

(\vec{r}- (\hat{i}+2\hat{j}-4\hat{k})).(-9\hat{i}+8\hat{j}-\hat{k}) = 0
\vec{r}.(9\hat{i}+8\hat{j}-\hat{k})=11

The cartesian equation of the plane is

-9x + 8y – z = 11

Now the distance of the point(9, -8, -10) from the plane is 

D = |\frac{(-9)(9)+(8)(-8)-(-10)-11}{\sqrt{9^2+8^2+1^2}}|

D = √146

Question 19. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line \frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}.

Solution:

When the plane passes through points (3, 4, 1), then the equation of the plane is 

a(x – 3) + b(y – 4) + c(z – 1)=0           ….(1)

When this plane passes through points (0, 1, 0), then the equation of the plane is

a(0 – 3) + b(1 – 4) + c(0 – 1) = 0 

⇒ 3a + 3b + 3c = 0          …..(2)

Also, given that the plane(i.e., eq(1)) is parallel to line. 

So the normal of the plane(i.e., eq(1)) is perpendicular to the line so,  

2a + 7b + 5c = 0      …..(3)

Now, on solving eq(1), (2), and (3), we get

8x – 13y + 15z + 13 = 0 is the required equation.

Question 20. Find the coordinates of the point where the line \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}      and intersects the plane x – y + z – 5 = 0. Also find the angle between the line and the plane.

Solution:

Given that the equation of line is \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda

⇒ x = 3λ + 2, y = 4λ – 1, z = 2λ + 2      ….(1)

As we know that (x, y, z) intersect the plane x – y + z – 5 = 0,

So, 

3λ + 2 – (4λ – 1) + 2λ + 2 – 5 = 0

⇒ λ = 0

Now, put this value in eq(1), we get

x = 2, y = -1, z = 2

The angle between the line and the plane is

sin\theta=\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}

Here, \vec{b} = 3\hat{i}+4\hat{j}+2\hat{k}

\vec{n} = \hat{i}-\hat{j}+\hat{k}

⇒ sin\theta=\frac{(3\hat{i}+4\hat{j}+2\hat{k}).(\hat{i}-\hat{j}+\hat{k})}{|3\hat{i}+4\hat{j}+2\hat{k}||\hat{i}-\hat{j}+\hat{k}|}

sin\theta=\frac{3-4+2}{\sqrt{9+16+4}\sqrt{1+1+1}}

⇒ \theta=sin^{-1}(\frac{1}{\sqrt{87}})

Question 21. Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane \vec{r}.(\hat{i}+2\hat{j}-5\hat{k})+9=0      .

Solution:

Let us assume that the direction ratios are a, b, c 

Given that the line passes through (1, 2, 3), so the equation of the line is 

\frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}                 ….(1)

And the line is perpendicular to the plane \vec{r}.(\hat{i}+2\hat{j}-5\hat{k})+9=0

So, the line is parallel to the normal of the plane.

Now, the direction ratios are proportional to those of the given plane.

⇒ a = λ, b = 2λ, c = -5λ

Put these values in eq(1), we get

\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-5}

So, the vector form is 

\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(\hat{i}+2\hat{j}-5\hat{k})      is the required equation.

Question 22. Find the angle between the line \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}   and the plane 10x + 2y – 11z = 3.

Solution:

Given that the equation of the line is \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}   and the equation of the plane is 10x + 2y – 11z = 3

So, \vec{b} = 2\hat{i}+3\hat{j}+6\hat{k}

\vec{n} = 10\hat{i}+2\hat{j}-11\hat{k}

As we know that the angle between a line and a plane is 

sin\theta=\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}

\frac{(2\hat{i}+3\hat{j}+6\hat{k}).(10\hat{i}+2\hat{j}-11\hat{k})}{|2\hat{i}+3\hat{j}+6\hat{k}||10\hat{i}+2\hat{j}-11\hat{k})|}

 ⇒ θ sin−1(−8/21​)

Question 23. Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \vec{r}.(\hat{i}-\hat{j}+2\hat{k})=5   and \vec{r}.(3\hat{i}+\hat{j}+\hat{k})=6.

Solution:

Let us assume, ab, b, c are the direction ratios of the required line.

Given that the line is passes through (1, 2, 3). So the equation of the line is 

\frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z - 3}{c}                      …(1)

Also given that eq(1) is parallel to the planes \vec{r}.(\hat{i}-\hat{j}+2\hat{k})= 5    and \vec{r}.(3\hat{i}+\hat{j}+2\hat{k})=6     .

So, a – b + 2c = 0          …(2)

3a + b + z = 0              …(3)

Now on solving eq(2) and (3), we get

\frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3}
\frac{a}{- 3} = \frac{b}{5} = \frac{c}{4} = \lambda

=> a = -3λ, b = 5λ, c = 4λ

Now put these values in eq(1), we get

\frac{x - 1}{- 3} = \frac{y - 2}{5} = \frac{z - 3}{4}     which is the cartesian form of the required line.

Question 24. Find the value of λ such that the line \frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}    is perpendicular to the plane 3x − y − 2z = 7.

Solution:

Given that the equation of line is \frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}   and

the equation of the plane is 3x − y − 2z = 7 and the line is perpendicular to the plane

So, the direction ratios of the given line are proportional to 6, λ ,-4.

and the direction ratios of the plane are 3, -1, -2.

Thus the line is parallel to the given plane, the line is perpendicular 

to the normal of the given plane. So, 

⇒ (6)(3) + (-1)(-4) + (-2 )(11) = 0

⇒ λ = 26

Question 25. Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}.

Solution:

The general equation of the plane passing through the point (−1, 2, 0) is

a(x+1) + b(y-2) + c( z – 0) = 0             ….(1) 

This plane passes through the point (2, 2,−1), we get

a(2 + 1) + b(2 – 2) + c( -1 – 0) = 0

⇒ 3a – c = 0             ….(2)

Now, a, b, c are the direction ratio of the normal to the plane (1) and 

the normal is perpendicular to the line, so

a + 2b + c = 0             ….(3)  

Now on solving eq(2) and (3), we get

\frac{a}{2} = \frac{b}{-4} = \frac{c}{6}

a = λ, b = -2 λ, c = 3λ

Now put all these values in eq(1), we get

λ(x + 1) – 2λ(y – 2) + 3λ(z – 0) = 0

x + 2y + 3z = 3

Thus, the equation of the required plane is x + 2y + 3z = 3.

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.