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

**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 (x

_{1},y_{1},z_{1}) is given bya(x−x

_{1}) + b(y−y_{1}) + c(z−z_{1}) = 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 (x

_{1},y_{1},z_{1}) is given bya(x−x

_{1}) + b(y−y_{1}) + c(z−z_{1}) = 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 a

_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0, we get3a − 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 r

_{1}= (−3,−2k,2) and r_{2}= (k,1,5).Since the lines are perpendicular, we have

(−3,−2k,2).(k,1,5) = 0

⇒ 3k + 2k − 10 = 0

⇒ 5k = 10

⇒ k = 2Now, 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.

## Please

Loginto comment...