**Chapter 6 Line and Plane Miscellaneous Exercise 6B**

## Chapter 6 Line and Plane Miscellaneous Exercise 6B

**Question 1.**

Solution:

Solution:

**Question 2.The vector equation of line 2x – 1 = 3y + 2 = z – 2 isSolution:**

**Question 3.The direction ratios of the line which is perpendicular to the two lines **

(A) 4, 5, 7

(B) 4, -5, 7

(C) 4, -5, -7

(D) -4, 5, 8

Solution:

(A) 4, 5, 7

**Question 4.**

**Question 5.**

**Question 6.**
(A) k = 1 or -1

(B) k = 0 or -3

(C) k = + 3

(D) k = 0 or -1

Solution:

Solution:

(B ) k = 0 or -3

**Question 7.**

(A) perpendicular

(B) inrersecting

(C) skew

(D) coincident
**Solution:**

(B) inrersecting

**Question 8.Equation of X-axis is**

(A) x = y = z

(B) y = z

(C) y = 0, z = 0

(D) x = 0, y = 0

**Solution:**

(C) y = 0, z = 0

**Question 9.The angle between the lines 2x = 3y = -z and 6x = -y = -4z is**

(A ) 45º

(B ) 30º

(C ) 0º

(D ) 90º

**Solution:**

(D ) 90º

**Question 10.The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are**

(A ) 2, 1, 6

(B ) 2, 1, -6

(C ) 2, -1, 6

(D ) -2, 1, 6

**Solution:**

(B ) 2, 1, -6

**Question 11.**

**Solution:**

(A ) 14

**Question 12.**

**Question 13.Solution:**

**Question 14.The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is**

(A ) x + y + z =1

(B ) x + y + z = 2

(C ) x + y + z = 3

(D ) x + y + z = 4

**Solution:**

(D ) x + y + z = 4

**Question 15.Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is**

(A ) 0º

(B ) 30º

(C ) 45º

(D ) 90º

**Solution:**

(A ) 0º

**Question 16.The direction cosines of the normal to the plane 2x – y + 2z = 3 areSolution:**

**Question 17.The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is :**

(A ) 3x + 2z – 1 = 0

(B ) 3x – 2z = 1

(C ) 3x + 2z + 1 = 0

(D ) 3x + 2z = 2

**Solution:**

(B ) 3x – 2z = 1

**Question 18.**

(A ) 17x – 47y – 24z + 172 = 0

(B ) 17x + 47y – 24z + 172 = 0

(C ) 17x + 47y + 24z +172 = 0

(D ) 17x – 47y + 24z + 172 = 0

**Solution:**

(A ) 17x – 47y – 24z + 172 = 0

**Question 19.**

(A ) 5

(B ) 3

(C ) 2

(D ) -5

**Solution:**

(A ) 5

**Question 20.The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is**

(A ) 4x + y + 5z = 14

(B ) 4x – 2y – 5z = 45

(C ) x – 2y – 5z = 10

(D ) 4x + y + 6z = 11

**Solution:**

(B ) 4x – 2y – 5z = 45

**II. Solve the following :Question 1.**

**Question 2.Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0Solution:**

The distance of the point (x

_{1}, y

_{1}, z

_{1}) from the plane ax + by + cz + d is

∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is

= 1units.

**Question 3.Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.Solution:**

The equation of the plane is 2x + 3y + 6z = 49

Dividing each term by

and length of perpendicular from origin to the plane is p = 7.

the coordinates of the foot of the perpendicular from the origin to the plane are

(lp, ∓, np)i.e.(2, 3, 6)

**Question 4.**

This is the normal form of the equation of plane.

**Question 5.Find the vector equation of the plane passing through the points A(1, -2, 1), B (2, -1, -3) and C (0, 1, 5).Solution:**

The vector equation of the plane passing through three non-collinear points

**Question 6.Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.Solution:**

The Cartesian equation of the plane passing through (x

_{1}, y

_{1}, z

_{1}), the direction ratios of whose normal are a, b, c, is

a(x – x

_{1}) + b(y – y

_{1}) + c(z – z

_{1}) = 0

∴ the cartesian equation of the required plane is

o(x + 1) + 2(y + 2) + 5(z – 3) = 0

i.e. 0 + 2y – 4 + 10z – 15 = 0

i.e. y + 2 = 0.

**Question 7.Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane **

Solution:

∴ its cartesian equation is

6x + 8y + 7z = p …(1)

A (7, 8, 6) lies on it and hence satisfies its equation

∴ (6)(7) + (8)(8) + (7)(6) = p

i.e., p = 42 + 64 + 42 = 148.

∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

**Question 8.The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). Find the vector equation of the plane.Solution:**

**Question 9.**

The required plane makes intercepts 1, 1, 1 on the coordinate axes.

∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0, , 1)

**Question 10.**

= (-2)(-4) + (7)(-1) + (5)(4)

= 8 – 7 + 8

= 35

∴ From (1), the vector equation of the required plane is

**Question 11.**

**Question 12.Find the vector equations of planes which pass through A(1, 2, 3), B (3, 2, 1) and make equal intercepts on the co-ordinates axes.Question is modifiedFind the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.Solution:**

Case 1 : Let all the intercepts be 0.

Then the plane passes through the origin.

Then the cartesian equation of the plane is

ax + by + cz = 0 …..(1)

(1, 2, 3) and (3, 2, 1) lie on the plane.

∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ a, b, c are proportional to 1, -2, 1

∴ from (1), the required cartesian equation is x – 2y + z = 0.

Case 2 : Let the plane make non zero intercept p on each axis.

i.e. x + y + z = p …(2)

Since this plane pass through (1, 2, 3) and (3, 2, 1)

∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p

∴ p = 6

∴ from (2), the required cartesian equation is

x + y + z = 6

Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

**Question 13.Find the vector equation of the plane which makes equal non-zero intercepts on the co-ordinates axes and passes through (1, 1, 1).Solution:**

Case 1 : Let all the intercepts be 0.

Then the plane passes through the origin.

Then the vector equation of the plane is ax + by + cz …(1)

(1, 1, 1) lie on the plane.

∴ 1a + 1b + 1c = 0

∴ from (1), the required cartesian equation is x – y + z = 0

Case 2 : Let he plane make non zero intercept p on each axis.

Since this plane pass through (1, 1, 1)

∴ 1 + 1 + 1 = p

∴ p = 3

**Question 14.Solution:**

The acute angle between the planes

= (1)(2) + (1)(1) + (2)(1)

= 2 + 1 + 2

= 5

Also,

**Question 15.**

= (2)(2) + (3)(-1) + (-6)(1)

= 4 – 3 – 6

= -5

**Question 16.Solution:**

**Question 17.**

= (3)(2) + (3)(3) + (1)(-6)

= 6 + 9 – 6

= 9

**Question 18.Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.Solution:**

= 19units.

**Question 19.**

**Question 20.Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B( 4, 3, -2) at right angle.Solution:**

**Question 21.Show thatlines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.Solution:**

Given lines are x = y, z = 0 and x + y = 0, z = 0.

It is clear that (0, 0, 0) satisfies both the equations.

∴ the lines intersect at O whose position vector is 0¯¯¯

Since z = 0 for both the lines, both the lines lie in XY- plane.

Hence, we have to find equation of XY-plane.

Z-axis is perpendicular to XY-plane.