Class 12 RD Sharma Solutions - Chapter 29 The Plane - Exercise 29.13

Question 1. Show that \vec{r} = (2\hat{j}-3\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})      and \vec{r}=(2\hat{i}+6\hat{j}+3\hat{k})+μ(2\hat{i}+3\hat{j}+4\hat{k})      are coplanar. Also, find the equation of the plane containing them.

Solution:

We know, the lines \vec{r} = \vec{a_1}+λ\vec{b_1}      and \vec{r} = \vec{a_2}+λ\vec{b_2}      are coplanar if:

\vec{a_1}.(\vec{b_1}×\vec{b_2}) = \vec{a_2}.(\vec{b_1}×\vec{b_2})

⇒ \vec{b_1}×\vec{b_2}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\2&3&4\\\end{array}\right|

⇒ \vec{b_1}×\vec{b_2}=-\hat{i}+2\hat{j}-\hat{k}

⇒ \vec{a_1}.(\vec{b_1}×\vec{b_2})=(2\hat{j}-3\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

⇒ \vec{a_1}.(\vec{b_1}×\vec{b_2}) = 0 + 4 + 3 = 7 

⇒ \vec{a_2}.(\vec{b_1}×\vec{b_2})=(2\hat{i}+6\hat{j}+3\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

⇒ \vec{a_2}.(\vec{b_1}×\vec{b_2})=-2+12-3=7 

Since, \vec{a_1}.(\vec{b_1}×\vec{b_2}) = \vec{a_2}.(\vec{b_1}×\vec{b_2})    , the lines are coplanar.

Equation of the plane containing them: \vec{r}.(\vec{b_1}×\vec{b_2})=\vec{a_1}.(\vec{b_1}×\vec{b_2})

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

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

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

Question 2. Show that the lines \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}     and \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}     are coplanar. Also, find the equation of the plane containing them.

Solution:

We know the lines \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}     and \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}     are coplanar if,

\left|\begin{array}{cc}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0

So, =\left|\begin{array}{cc}0+1&7-3&-7+2\\-3&2&1\\1&-3&2\\\end{array}\right|

= \left|\begin{array}{cc}1&4&-5\\-3&2&1\\1&-3&2\\\end{array}\right|

= 1(4 + 3) − 4(−6 − 1) − 5(9 − 2)

= 7 + 28 − 35 

= 0.

So the lines are coplanar.

Equation of the plane: = \left|\begin{array}{cc}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0

 = \left|\begin{array}{cc}x+1&y-3&z+2\\-3&2&1\\1&-3&2\\\end{array}\right|=0

⇒ 7x + 7y + 7z = 0.

Question 3. Find the equation of the plane containing the line \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}     and the point (0,7,-7) and show that the line \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}     also lies in the same plane.

Solution:

We know the equation of a plane passing through a point (x₁,y₁,z₁) is given by

a(x−x₁) + b(y−y₁) + c(z−z₁) = 0            ........(1)

Since the required plane passes through (0,7,-7), the equation becomes

ax + b(y − 7) + c(z + 7) = 0                .......(2)

It also contains \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}     and point is (−1,3,−2).

a(−1) + b(3 − 7) + c(−2 + 7) = 0

⇒ −a − 4b + 5c = 0

Also, −3a + 2b + c = 0

Solving the equations, we get x + y + z = 0

So, \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}     lies on the plane x + y + z = 0.

Question 4. Find the equation of the plane which contains two parallel lines \frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-2}{5}     and \frac{x-3}{1}=\frac{y+2}{-4}=\frac{z}{5}.

Solution:

We know the equation of a plane passing through a point (x₁,y₁,z₁) is given by

a(x−x₁) + b(y−y₁) + c(z−z₁) = 0            ........(1)

The required plane passes through (4,3,2). Hence,

a(x − 4) + b(y − 3) + c(z − 2) = 0

It also passes through (3,-2,0). Hence,

a(3 − 4) + b(−2 − 3) + c(0 − 2) = 0

⇒ a + 5b + 2c = 0                .......(2)

Also, a − 4b + 5c = 0             ........(3)

Solving (2) and (3)  by cross multiplication, we get the equation of the plane as:

⇒ 11x − y − 3z − 35 = 0.

Question 5. Show that the lines \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}     and 3x − 2y + 5 = 0 = 2x + 3y + 4z − 4  intersect. Find the equation of the plane.

Solution:

Using a₁a₂ + b₁b₂ + c₁c₂ = 0, we get

3a − 2b + c = 0                ....(1)

Also, 2a + 3b + 4c = 0.     ....(2)

Solving (1) and (2) by cross multiplication, we have

\frac{a}{(-2)(4)-(3)(1)}=\frac{b}{(2)(1)-(3)(4)}=\frac{c}{(3)(3)-(-2)(2)}

⇒ \frac{a}{-11}=\frac{b}{-10}=\frac{c}{13}

Hence, the equation of the plane is 45x − 17y + 25z + 53 = 0

and the point of intersection is (2,4,−3).

Question 6. Show that the plane whose vector equation is \vec{r}.(\hat{i}+2\hat{j}-\hat{k})=3     contains the line whose vector equation is \vec{r}=\hat{i}+\hat{j}+λ(2\hat{i}+\hat{j}+4\hat{k}).       

Solution:

Here, \vec{b}.\vec{n}=(2\hat{i}+\hat{j}+4\hat{k}).(\hat{i}+2\hat{j}-\hat{k})

= 2(1) +1(2) + 4(−1)

⇒ \vec{b}.\vec{n} = 0

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

= 1(1) + 1(2) + 0(−1)

= 3 

⇒ \vec{a}.\vec{n}=d 

Hence, the given line lies on the plane.

Question 7. Find the equation of the plane determined by the intersection of the lines \frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}     and \frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}.

Solution:

Let the plane be ax + by + cz + d = 0

Since the plane passes through the intersection of the given lines, normal of the plane is perpendicular to the two lines.

⇒ 3a − 2b + 6c = 0

and, a − 3b + 2c = 0

Using cross multiplication, we have

\frac{a}{-4+18}=\frac{b}{6-6}=\frac{c}{-9+2}

⇒ \frac{a}{2}=\frac{b}{0}=\frac{c}{-1}

Question 8. Find the vector equation of the plane passing through the points (3,4,2) and (7,0,6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \vec{r}=\hat{i}+3\hat{j}-2\hat{k}+λ(\hat{i}-\hat{j}+\hat{k})

Solution:

Let the equation of the plane be \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

Since the plane passes through (3,4,2) and (7,0,6), we have

\frac{3}{a}+\frac{4}{b}+\frac{2}{c}=1     and \frac{7}{a}+\frac{0}{b}+\frac{6}{c}=1

Since the required plane is perpendicular to 2x − 5y − 15 = 0, we have,

\frac{2}{a}+\frac{(-5)}{b}+\frac{0}{c}=1

⇒ b = 2.5a

Substituting the value of b in the above equations we have, 

\frac{3}{a}+\frac{4}{2.5a}+\frac{2}{c}=1     and \frac{7}{a}+\frac{6}{c}=1

Solving the above equations, we have

a = 17/5, b = 17/2 and c = −17/3.

Substituting the values in the equation of plane, we obtain

5x + 2y − 3z = 17.

Vector equation of the plane becomes: \vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17    .

Question 9. If the lines \frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}     and \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}     are perpendicular, find the values of k and also the equation the plane containing these lines.

Solution:

The direction ratios of the two lines are r₁ = (−3,−2k,2) and r₂ = (k,1,5).

Since the lines are perpendicular, we have

 (−3,−2k,2).(k,1,5) = 0

⇒ 3k + 2k − 10 = 0

⇒ 5k = 10

⇒ k = 2

Now, equation of the plane containing the lines is: \left|\begin{array}{cc}x&y&z\\-3&-4&2\\2&1&5\\\end{array}\right|=0

⇒ −22x + 19y + 5z + 31 = 0.

Question 10. Find the coordinates of the point where the line \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}     intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.

Solution:

Any point on the given line is of the form (3k + 2, 4k − 1, 2k + 2).

We have, (3k + 2) − (4k − 1) + (2k + 2) − 5 = 0

⇒ k = 0.

Thus, the coordinates of the point become (2,−1,2).

Let v be the angle between the line and the plane. Then, 

sinv=\frac{al+bm+cn}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}

Here, l = 3, m = 4, n =2, a =1, b = −1, c = 1.

Hence, sinv=\frac{1(3)+4(-1)+1(2)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{3^2+4^2+2^2}}

⇒ sinv=\frac{1}{\sqrt{3}\sqrt{29}}

⇒ v=sin^{-1}[\frac{1}{\sqrt3\sqrt29}]

Question 11. Find the vector equation of the plane passing through three points with position vectors \hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k},\hat{i}+2\hat{j}+\hat{k}.

Solution:

Let  A, B and C be the three given vectors respectively.

\vec{AB}=(2\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}-2\hat{k})

⇒ \vec{AB}=\hat{i}-2\hat{j}+3\hat{k}

and, \vec{AC}=\hat{j}+3\hat{k}

Now, \vec{AB}.\vec{AC} = \left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\0&1&3\\\end{array}\right|

⇒ \vec{n}=-9\hat{i}-3\hat{j}+\hat{k}

Equation of the plane is: \vec{r}.(9\hat{i}+3\hat{j}-\hat{k})=14.

Coordinates of the points are (1,1,−2).

Question 12. Show that the lines \frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}     and \frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}     are coplanar. 

Solution:

We know the lines  \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}     and \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}     are coplanar if,

\left|\begin{array}{cc}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0

or, \left|\begin{array}{cc}8-5&4-7&5+3\\4&4&-5\\7&1&3\\\end{array}\right|

=  \left|\begin{array}{cc}3&-3&8\\4&4&-5\\7&1&3\\\end{array}\right|

= 3(12 + 5) + 3(12 + 35) + 8(4 − 28)

= 0.

Hence the lines are coplanar.

Question 13. Find the equation of the plane which passes through the point (3,2,0) and contains the line \frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}.

Solution:

Required equation of the plane passing through (3,2,0) is:

a(x − 3) + b(y − 2) + cz = 0            ......(1)

Since the plane also passes through the given line, we have

4b + 4c = 0               ......(2)

Also, the plane will be parallel so,

a + 5b + 4c = 0        ......(3)

Solving (2) and (3), we have

\frac{a}{16-20}=\frac{b}{4-0}=\frac{c}{0-4}=z

⇒ -\frac{a}{4}=\frac{b}{4}=-\frac{c}{4}=z

⇒ a = −z, b = z and c = −z

Putting the values in (1), we have

x − y + z − 1 = 0.

Summary

Exercise 29.13 likely focuses on advanced problems related to planes in three-dimensional coordinate geometry. It may cover topics such as finding equations of planes under various conditions, calculating distances between points and planes, determining angles between planes, and analyzing the relationships between multiple planes. The exercise probably emphasizes practical applications and complex problem-solving skills involving planes.