Chapter 11 Three Dimensional Geometry Ex 11.3 Solutions
Question - 1 : - In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
Answer - 1 : -
(a) z = 2
Given:
The equation of theplane, z = 2 or 0x + 0y + z = 2 …. (1)
Direction ratio of thenormal (0, 0, 1)
By using the formula,
√[(0)2 +(0)2 + (1)2] = √1
= 1
Now,
Divide both the sidesof equation (1) by 1, we get
0x/(1) + 0y/(1) + z/1= 2
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0, 0, 1
Distance (d) from theorigin is 2 units
(b) x + y + z = 1
Given:
The equation of theplane, x + y + z = 1…. (1)
Direction ratio of thenormal (1, 1, 1)
By using the formula,
√[(1)2 +(1)2 + (1)2] = √3
Now,
Divide both the sidesof equation (1) by √3, weget
x/(√3) + y/(√3) + z/(√3) = 1/√3
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 1/√3, 1/√3, 1/√3
Distance (d) from theorigin is 1/√3 units
(c) 2x + 3y – z = 5
Given:
The equation of theplane, 2x + 3y – z = 5…. (1)
Direction ratio of thenormal (2, 3, -1)
By using the formula,
√[(2)2 +(3)2 + (-1)2] = √14
Now,
Divide both the sidesof equation (1) by √14, weget
2x/(√14) + 3y/(√14) – z/(√14) = 5/√14
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 2/√14, 3/√14, -1/√14
Distance (d) from theorigin is 5/√14 units
(d) 5y + 8 = 0
Given:
The equation of theplane, 5y + 8 = 0
-5y = 8 or
0x – 5y + 0z = 8…. (1)
Direction ratio of thenormal (0, -5, 0)
By using the formula,
√[(0)2 +(-5)2 + (0)2] = √25
= 5
Now,
Divide both the sidesof equation (1) by 5, we get
0x/(5) – 5y/(5) –0z/(5) = 8/5
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0, -1, 0
Distance (d) from theorigin is 8/5 units
Question - 2 : - Find the vector equation of a plane which is at a distance of 7 units fromthe origin and normal to the vector
Answer - 2 : -
Question - 3 : - Find the Cartesian equation of the following planes:
(a)
Answer - 3 : - (a)
Given:
The equation of theplane.
(b)
(c)
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Answer - 4 : -
(a) 2x + 3y + 4z – 12 = 0
Let the coordinate ofthe foot of ⊥ P from theorigin to the given plane be P(x, y, z).
2x + 3y + 4z = 12 ….(1)
Direction ratio are(2, 3, 4)
√[(2)2 +(3)2 + (4)2] = √(4 + 9 + 16)
= √29
Now,
Divide both the sidesof equation (1) by √29, weget
2x/(√29) + 3y/(√29) + 4z/(√29) =12/√29
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 2/√29, 3/√29, 4/√29
Coordinate of the foot(ld, md, nd) =
= [(2/√29) (12/√29), (3/√29) (12/√29), (4/√29) (12/√29)]
= 24/29, 36/29, 48/29
(b) 3y + 4z – 6 = 0
Let the coordinate ofthe foot of ⊥ P from theorigin to the given plane be P(x, y, z).
0x + 3y + 4z = 6 ….(1)
Direction ratio are(0, 3, 4)
√[(0)2 +(3)2 + (4)2] = √(0 + 9 + 16)
= √25
= 5
Now,
Divide both the sidesof equation (1) by 5, we get
0x/(5) + 3y/(5) +4z/(5) = 6/5
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0/5, 3/5, 4/5
Coordinate of the foot(ld, md, nd) =
= [(0/5) (6/5), (3/5)(6/5), (4/5) (6/5)]
= 0, 18/25, 24/25
(c) x + y + z = 1
Let the coordinate ofthe foot of ⊥ P from theorigin to the given plane be P(x, y, z).
x + y + z = 1 …. (1)
Direction ratio are(1, 1, 1)
√[(1)2 +(1)2 + (1)2] = √(1 + 1 + 1)
= √3
Now,
Divide both the sidesof equation (1) by √3, weget
1x/(√3) + 1y/(√3) + 1z/(√3) = 1/√3
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 1/√3, 1/√3, 1/√3
Coordinate of the foot(ld, md, nd) =
= [(1/√3) (1/√3), (1/√3) (1/√3), (1/√3) (1/√3)]
= 1/3, 1/3, 1/3
(d) 5y + 8 = 0
Let the coordinate ofthe foot of ⊥ P from theorigin to the given plane be P(x, y, z).
0x – 5y + 0z = 8 ….(1)
Direction ratio are(0, -5, 0)
√[(0)2 +(-5)2 + (0)2] = √(0 + 25 + 0)
= √25
= 5
Now,
Divide both the sidesof equation (1) by 5, we get
0x/(5) – 5y/(5) +0z/(5) = 8/5
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0, -1, 0
Coordinate of the foot(ld, md, nd) =
= [(0/5) (8/5), (-5/5)(8/5), (0/5) (8/5)]
= 0, -8/5, 0
Question - 5 : - Find the vector and Cartesian equations of the planes
(a) that passes through the point (1, 0, –2) and the normal to the plane is
(b) that passes through the point (1,4, 6) and the normal vector to the plane is
Answer - 5 : -
x – 1 – 2y + 8 + z – 6= 0
x – 2y + z + 1 = 0
x – 2y + z = -1
∴ The requiredCartesian equation of the plane is x – 2y + z = -1
x – 1 – 2y + 8 + z – 6= 0
x – 2y + z + 1 = 0
x – 2y + z = -1
∴ The requiredCartesian equation of the plane is x – 2y + z = -1
Find the equations of the planes that passes through three points.
(a) (1, 1, –1), (6, 4, –5), (–4, –2, 3)
(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)
Answer - 6 : -
Given:
The points are (1, 1,-1), (6, 4, -5), (-4, -2, 3).
Let,
= 1(12 – 10) – 1(18 –20) -1 (-12 + 16)
= 2 + 2 – 4
= 0
Since, the value ofdeterminant is 0.
∴ The points arecollinear as there will be infinite planes passing through the given 3 points.
(b) (1, 1, 0), (1, 2, 1),(–2, 2, –1)
Question - 7 : - Find the intercepts cut off by the plane 2x + y – z = 5.
Answer - 7 : -
Given:
The plane 2x + y – z =5
Let us express theequation of the plane in intercept form
x/a + y/b + z/c = 1
Where a, b, c are theintercepts cut-off by the plane at x, y and z axes respectively.
2x + y – z =5 …. (1)
Now divide both thesides of equation (1) by 5, we get
2x/5 + y/5 – z/5 = 5/5
2x/5 + y/5 – z/5 = 1
x/(5/2) + y/5 + z/(-5)= 1
Here, a = 5/2, b = 5and c = -5
∴ The interceptscut-off by the plane are 5/2, 5 and -5.
Question - 8 : - Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Answer - 8 : -
We know that theequation of the plane ZOX is y = 0
So, the equation ofplane parallel to ZOX is of the form, y = a
Since the y-interceptof the plane is 3, a = 3
∴ The required equationof the plane is y = 3
Question - 9 : - Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
Answer - 9 : -
Given:
Equation of the planepasses through the intersection of the plane is given by
(3x – y + 2z – 4)+ λ (x + y + z – 2) = 0 and the plane passes through the points (2, 2, 1).
So, (3 × 2 – 2 + 2 × 1– 4) + λ (2 + 2 + 1 – 2) = 0
2 + 3λ = 0
3λ = -2
λ = -2/3 …. (1)
Upon simplification,the required equation of the plane is given as
(3x – y + 2z – 4) –2/3 (x + y + z – 2) = 0
(9x – 3y + 6z – 12 –2x – 2y – 2z + 4)/3 = 0
7x – 5y + 4z – 8 = 0
∴ The required equationof the plane is 7x – 5y + 4z – 8 = 0
Question - 10 : - Find the vector equation of the plane passing through the intersection of the planes 
and through the point (2, 1, 3).
Answer - 10 : -
The equation of anyplane through the intersection of the planes given in equations (1) and (2) isgiven by,
