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

  • Last Updated : 20 May, 2021

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 multipliacation, 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 olane 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.


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