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

• Last Updated : 18 Feb, 2022

### Question 1. Show that and are coplanar. Also, find the equation of the plane containing them.

Solution:

We know, the lines and are coplanar if:       Since, , the lines are coplanar.

Equation of the plane containing them:    ### Question 2. Show that the lines and are coplanar. Also, find the equation of the plane containing them.

Solution:

We know the lines and are coplanar if, So,  = 1(4 + 3) − 4(−6 − 1) − 5(9 − 2)

= 7 + 28 − 35

= 0.

So the lines are coplanar.

Equation of the plane:  ⇒ 7x + 7y + 7z = 0.

### Question 3. Find the equation of the plane containing the line and the point (0,7,-7) and show that the line 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 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 lies on the plane x + y + z = 0.

### Question 4. Find the equation of the plane which contains two parallel lines and 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 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  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 contains the line whose vector equation is Solution:

Here, = 2(1) +1(2) + 4(−1) Now, = 1(1) + 1(2) + 0(−1)

= 3 Hence, the given line lies on the plane.

### Question 7. Find the equation of the plane determined by the intersection of the lines and 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 ⇒ ### 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 Solution:

Let the equation of the plane be Since the plane passes through (3,4,2) and (7,0,6), we have and Since the required plane is perpendicular to 2x − 5y − 15 = 0, we have, ⇒ b = 2.5a

Substituting the value of b in the above equations we have, and 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: .

### Question 9. If the lines and 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: ⇒ −22x + 19y + 5z + 31 = 0.

### Question 10. Find the coordinates of the point where the line 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, Here, l = 3, m = 4, n =2, a =1, b = −1, c = 1.

Hence, ⇒ ⇒ ### Question 11. Find the vector equation of the plane passing through three points with position vectors Solution:

Let  A, B and C be the three given vectors respectively.  and, Now, ⇒ Equation of the plane is: Coordinates of the points are (1,1,−2).

### Question 12. Show that the lines and are coplanar.

Solution:

We know the lines and are coplanar if, or,  = 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 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 ⇒ ⇒ a = −z, b = z and c = −z

Putting the values in (1), we have

x − y + z − 1 = 0.

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