# Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.11 | Set 1

**Question 1. Find the angle between the line and the plane .**

**Solution:**

Given that, the line is and the plane is .

So,

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

So,

⇒

⇒ 9/√87

**Question 2. Find the angle between the line and the plane 2x + y – z = 4.**

**Solution:**

The given line is parallel to the vector and the plane 2x + y – z = 4 is normal to the vector .

So, the angle between line and plane is

=

= 0

**Question 3. Find the angle between the line joining the points (3, -4, -2) and (12, 2, 0) and the plane 3x – y + z = 1.**

**Solution:**

According to the question, the line passes through A(3,- 4,- 2) and B(12, 2, 0).

So,

=

=

So, the line is parallel to the vector and the plane is normal to the vector

So, the angle between the line and the plane is,

=

=

Thus, .

**Question 4. The line is parallel to the plane . Find m.**

**Solution:**

Give that the equation of line is and the equation of plane is

So,

If a line is parallel to a plane, then the normal to the plane is perpendicular to the line.

⇒

⇒

⇒

⇒ 2m – 3m – 3 = 0

⇒ – m – 3 = 0

⇒ m = –3

**Question 5. Show that the line whose vector equation is is parallel to the plane whose vector . Also, find the distance between them.**

**Solution:**

Given that the plane passes through the point with the position vector and is parallel to the vector .

So, the normal vector and d = 7 .

= 1 + 3 – 4

= 4 – 4

= 0

As we know that is perpendicular to

So, the distance between the line and plane is

= 7/√3units.

**Question 6. Find the vector equation of the line through the origin which is perpendicular to the plane .**

**Solution:**

Given that the line is perpendicular to the plane

So, the line is parallel to the normal .

As we know that the equation of a line is pass through and parallel to is

⇒

⇒

**Question 7. Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to the x-axis.**

**Solution:**

Given that the equation of the plane through point (2, 3, -4) is

a(x − 2) + b(y − 3)+c(z + 4) = 0 …(1)

Since this plane passes through point (1, -1, 3).

⇒ a(1 − 2) + b( −1 − 3) + c( 3 + 4) = 0

⇒ − a − 4b + 7c = 0 …(2)

Equation(1) is parallel to x-axis and is perpendicular to the yz-plane whose equation is x = 0 or 1 . x + 0 . y + 0 . z = 0

⇒ a(1) + b(0) + c(0) = 0 …(3)

One solving eq(1), (2), and (3), we get

⇒ 7y + 4z – 5 = 0 is the required equation.

**Question 8. Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line **

**Solution:**

Given that the plane pass through point(0, 0, 0), so the equation of the plane is a(x − 0) + b(y − 0)+c(z + 0) = 0.

⇒ ax + by + cz = 0 …(1)

and the same plane passes through point (3, -1, 2). So, the equation of the plane is

3a – b + 2c = 0 …(2)

Equation(1) is parallel to the given line so,

a(1) + b(-4) + c (7) = 0 …(3)

On solving eq(1), (2), and (3), we get

⇒ x – 19y – 11z = 0

Hence, the required equation of the plane is x – 19y – 11z = 0

### Question 9. Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes and

**Solution:**

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

…(1)

Here, the given line is parallel to plane x – y + 2z = 5

So, a × 1 + b × – 1 + c × 2 = 0

⇒ a – b + 2c = 0 …(3)

Also, the given line is parallel to plane 3x + y + z = 6

So, a × 3 + b × 1 + c × 1 = 0

⇒ 3a + b + c = 0 …(4)

On solving the equation (3) and (4) we get,

⇒

∴ a = – 3k, b = 5k and c = 4k

Now, put the value in the equation(1), we get

Now, multiplying by k we get

The required equation is

⇒

Hence, the equation of the plane is

### Question 10. Prove that the line of section of the plane 5x + 2y – 4z + 2 = 0 and 2x + 8y + 2z – 1 = 0 is parallel to the plane 4x – 2y – 5z – 2 = 0.

**Solution:**

Let us consider a, b and c be the direction ratios.

So,

⇒ a + 4b + c = 0

and, 5a + 2b – 4c = 0

On solving the above two equations, we get

⇒

As we know that the line is parallel to plan a

_{2}x + b_{2}y + c_{2}z + d_{2}= 0, when a_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0So, the line with direction ratio(a, b, c) is parallel to plane 4x – 2y – 5z – 2 = 0

aa

_{1}+ bb_{1}+ cc_{1}= 02(4) + (-1)(-2) + (2)(-5) = 0

Hence proved that the line of section of the given plane is parallel to the given plane.

### Question 11. Find the vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x – y + 3z – 5 = 0.

**Solution:**

Let us consider a, b and c be the direction ratios.

Given that the equation of the line passing through the point (1, -1, 2)

So,

…..(1)

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

⇒

⇒ a = 2λ, b = -λ , c = 3λ

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

So, the line passes through a point whose position vector is and parallel to

So,

⇒is the required equation.

### Question 12. Find the equation of the plane through the points (2, 2, -1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

**Solution:**

Given that the equation of the plane through the point (2, 2, -1)

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

Also, the plane passes through points (3, 4, 2)

a(3 – 2) + b(4 – 2)+ c(2 + 1) = 0

or, a + 2b + 3c = 0 …(2)

As we know that the line is parallel to plan a

_{2}x + b_{2}y + c_{2}z + d_{2}= 0, when a_{1}a_{2}+ b_{1}b_{2}+ c_{1}c_{2}= 0According to the question, the plane(1) is parallel to the line whose direction ratios are 7, 0, 6

So, 7a + 0b + 6c = 0 …(3)

Now, on solving the equation (1), (2), and (3), we have,

⇒ 12x + 15y – 14z – 68 = 0 is the required equation.

### Question 13. Find the angle between the line and the plane 3x + 4y + z + 5 = 0.

**Solution:**

Given that the equation of line is and the equation of the plane is 3x + 4y + z + 5 = 0

So,

Angle between a line and a plane is

⇒

⇒

⇒

Thus,

## Please

Loginto comment...