Chapter 11 Three Dimensional Geometry Ex 11.3 Solutions
Question - 11 : - Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Answer - 11 : -
Let the equation ofthe plane that passes through the two-given planes
x + y + z = 1 and 2x +3y + 4z = 5 is
(x + y + z – 1)+ λ (2x + 3y + 4z – 5) = 0
(2λ + 1) x +(3λ + 1) y + (4λ + 1) z -1 – 5λ = 0…… (1)
So the direction ratioof the plane is (2λ + 1, 3λ + 1, 4λ + 1)
And direction ratio ofanother plane is (1, -1, 1)
Since, both the planesare ⊥
So by substituting ina1a2 + b1b2 + c1c2 =0
(2λ + 1 × 1)+ (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0
2λ + 1 –3λ – 1 + 4λ + 1 = 0
3λ + 1 = 0
λ = -1/3
Substitute the valueof λ in equation (1) we get,
x – z + 2 = 0
∴ The required equationof the plane is x – z + 2 = 0
Question - 12 : - Find the angle between the planes whose vector equations are
Answer - 12 : -
Question - 13 : - In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Answer - 13 : -
(a) 7x + 5y + 6z + 30 = 0and 3x – y – 10z + 4 = 0
Given:
The equation of thegiven planes are
7x + 5y + 6z + 30 = 0and 3x – y – 10z + 4 = 0
Two planes are ⊥ if the direction ratio of the normal tothe plane is
a1a2 +b1b2 + c1c2 = 0
21 – 5 – 60
-44 ≠ 0
Both the planes arenot ⊥ to each other.
Now, two planes are ||to each other if the direction ratio of the normal to the plane is
∴ The angle is cos-1 (2/5)
(b) 2x + y + 3z – 2 = 0and x – 2y + 5 = 0
Given:
The equation of thegiven planes are
2x + y + 3z – 2 = 0and x – 2y + 5 = 0
Two planes are ⊥ if the direction ratio of the normal tothe plane is
a1a2 +b1b2 + c1c2 = 0
2 × 1 + 1 × (-2) + 3 ×0
= 0
∴ The given planesare ⊥ to each other.
(c) 2x – 2y + 4z + 5 = 0and 3x – 3y + 6z – 1 = 0
Given:
The equation of thegiven planes are
2x – 2y + 4z + 5 =0and x – 2y + 5 = 0
We know that, twoplanes are ⊥ if the directionratio of the normal to the plane is
a1a2 +b1b2 + c1c2 = 0
6 + 6 + 24
36 ≠ 0
∴ Both the planesare not ⊥ to each other.
Now let us check, bothplanes are || to each other if the direction ratio of the normal to the planeis
∴ The given planes are|| to each other.
(d) 2x – 2y + 4z + 5 = 0and 3x – 3y + 6z – 1 = 0
Given:
The equation of thegiven planes are
2x – y + 3z – 1 = 0and 2x – y + 3z + 3 = 0
We know that, twoplanes are ⊥ if the directionratio of the normal to the plane is
a1a2 +b1b2 + c1c2 = 0
2 × 2 + (-1) × (-1) +3 × 3
14 ≠ 0
∴ Both the planesare not ⊥ to each other.
Now, let us check twoplanes are || to each other if the direction ratio of the normal to the planeis
∴ The given planes are|| to each other.
(e) 4x + 8y + z – 8 = 0and y + z – 4 = 0
Given:
The equation of thegiven planes are
4x + 8y + z – 8 = 0and y + z – 4 = 0
We know that, twoplanes are ⊥ if the directionratio of the normal to the plane is
a1a2 +b1b2 + c1c2 = 0
0 + 8 + 1
9 ≠ 0
∴ Both the planesare not ⊥ to each other.
Now let us check, twoplanes are || to each other if the direction ratio of the normal to the planeis
∴ Both the planesare not || to each other.
Now let us find theangle between them which is given as
∴ The angle is 45o.
Question - 14 : - In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3
(b) (3, -2, 1) 2x – y + 2z + 3 = 0
(c) (2, 3, -5) x + 2y – 2z = 9
(d) (-6, 0, 0) 2x – 3y + 6z – 2 = 0
Answer - 14 : -
(a) Point Plane
(0, 0, 0) 3x – 4y + 12z = 3
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (0, 0,0) and the plane is 3x – 4y + 12z = 3
= |3/√169|
= 3/13
∴ The distance is 3/13.
(b) Point Plane
(3, -2, 1) 2x – y + 2z+ 3 = 0
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (3, -2,1) and the plane is 2x – y + 2z + 3 = 0
= |13/√9|
= 13/3
∴ The distance is 13/3.
(c) Point Plane
(2, 3, -5) x + 2y – 2z= 9
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (2, 3,-5) and the plane is x + 2y – 2z = 9
= |9/√9|
= 9/3
= 3
∴ The distance is 3.
(d) Point Plane
(-6, 0, 0) 2x – 3y +6z – 2 = 0
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (-6, 0,0) and the plane is 2x – 3y + 6z – 2 = 0
= |14/√49|
= 14/7
= 2
∴ The distance is 2.