NCERT Solutions for Maths Class 12 Exercise 11.3
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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.
NCERT Solutions for Maths Class 12 Exercise 11.3 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.
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NCERT Solutions for Maths Class 12 Exercise 11.1
NCERT Solutions for Maths Class 12 Exercise 11.2
NCERT Solutions for Maths Class 12 Exercise 11.3
Maths Class
12 Ex 11.3 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
Solution:
(a) Direction ratios of the normal to the plane
are 0, 0, 1.
Here, a = 0, b = 0, c = 1
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i + 5j – 6k.
Solution:
Find the
Cartesian equation of the following planes.
Solution:
(a) Let r is
the position vector of any arbitrary point P(x, y, z) on the plane.
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
Solution:
(a) Let N(x1, y1, z1)
be the foot of the perpendicular from the origin to the plane 2x + 3y + 4z – 12
= 0.
∴ Direction ratios of the normal are 2,
3, 4.
Also, the direction ratios of ON are (x1, y1, z1)
Maths Class
12 Ex 11.3 Question 5.
Find the vector
and Cartesian equation of the planes
(a) that passes through the point (1, 0, –2) and the normal to the plane is i
+ j
– k.
(b) that passes through the point (1, 4, 6) and the normal vector to the plane
is i
– 2j
+ k.
Solution:
(a) Normal to the plane is i + j – k and passes through (1, 0,
–2)
Maths Class
12 Ex 11.3 Question 6.
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)
Solution:
(a) The plane passes through the points (1, 1, –1),
(6, 4, –5), (–4, –2, 3)
Let the equation of the plane passing through (1, 1, –1) be
Maths Class
12 Ex 11.3 Question 7.
Find the
intercepts cut off by the plane 2x + y – z = 5.
Solution:
Equation of the
plane is 2x + y – z = 5
Dividing by 5:
∴ The intercepts on the axes OX, OY, OZ
are 5/2, 5, –5, respectively.
Maths Class
12 Ex 11.3 Question 8.
Find the equation
of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Solution:
Any plane
parallel to ZOX plane is y = b, where b is the intercept on y-axis.
∴ b = 3.
Hence, equation of the required plane is y = 3.
Maths Class
12 Ex 11.3 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).
Solution:
Given planes are:
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0
Any plane through their intersection is
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
… (i)
Point (2, 2, 1) lies on it,
∴
3 × 2 – 2 + 2 × 1
– 4 + λ(2 + 2 + 1 – 2) = 0
Thus, λ = –2/3
Putting the value of λ in equation (i), we get the required equation is 7x – 5y
+ 4z – 8 = 0.
Maths Class
12 Ex 11.3 Question 10.
Find the vector
equation of the plane passing through the intersection of the planes r.(2i
+ 2j
– 3k)
= 7 and r.(2i + 5j + 3k) = 9 and through
the point (2, 1, 3).
Solution:
The equation of
the plane passing through the line of intersection of the planes
Maths Class
12 Ex 11.3 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.
Solution:
Given planes are
x + y + z – 1 = 0 … (i)
2x + 3y + 4z – 5 = 0 … (ii)
x – y + z = 0 … (iii)
Maths Class
12 Ex 11.3 Question 12.
Find the angle
between the planes whose vector equations are r.(2i + 2j – 3k) = 5 and r.(3i
– 3j
+ 5k)
= 3.
Solution:
The angle θ
between the given planes is
Maths Class
12 Ex 11.3 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 – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Solution:
(a) Direction ratios of the normal of the planes
7x + 5y + 6z + 30 = 0 are 7, 5, 6.
Direction ratios of the normal of the plane 3x – y – 10z + 4 = 0 are 3, –1, –10.
The plane 7x + 5y + 6z + 30 = 0 … (i)
3x – y – 10z + y = 0
… (ii)
Maths Class
12 Ex 11.3 Question 14.
In the following
cases, find the distance of each of the given points from the corresponding given
plane.
(a) Point (0, 0, 0); Plane 3x – 4y + 12z = 3
(b) Point (3, –2, 1); Plane 2x – y + 2z + 3 = 0
(c) Point (2, 3, –5); Plane x + 2y – 2z = 9
(d) Point (–6, 0, 0); Plane 2x – 3y + 6z – 2 = 0
Solution:
(a) Given plane: 3x – 4y + 12z – 3 = 0
