NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2.
- Introduction to Three Dimensional Geometry Class 11 Ex 12.1
- Introduction to Three Dimensional Geometry Class 11 Ex 12.3
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 12 |
| Chapter Name | Introduction to Three Dimensional Geometry |
| Exercise | Ex 12.2 |
| Number of Questions Solved | 5 |
| Category | NCERT Solutions |
NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2
Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is
=
(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is
![PQ=\sqrt { \left[ 2-\left( -3 \right) \right] ^{ 2 }+\left( 4-7 \right) ^{ 2 }\left( -1-2 \right) ^{ 2 } }](http://s0.wp.com/latex.php?latex=PQ%3D%5Csqrt+%7B+%5Cleft%5B+2-%5Cleft%28+-3+%5Cright%29+%5Cright%5D+%5E%7B+2+%7D%2B%5Cleft%28+4-7+%5Cright%29+%5E%7B+2+%7D%5Cleft%28+-1-2+%5Cright%29+%5E%7B+2+%7D+%7D+&bg=ffffff&fg=000&s=0 "PQ=\sqrt { \left[ 2-\left( -3 \right) \right] ^{ 2 }+\left( 4-7 \right) ^{ 2 }\left( -1-2 \right) ^{ 2 } }")
(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is
![PQ=\sqrt { \left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( -3-3 \right) ^{ 2 }\left[ 4-\left( -4 \right) \right] ^{ 2 } }](http://s0.wp.com/latex.php?latex=PQ%3D%5Csqrt+%7B+%5Cleft%5B+1-%5Cleft%28+-1+%5Cright%29+%5Cright%5D+%5E%7B+2+%7D%2B%5Cleft%28+-3-3+%5Cright%29+%5E%7B+2+%7D%5Cleft%5B+4-%5Cleft%28+-4+%5Cright%29+%5Cright%5D+%5E%7B+2+%7D+%7D+&bg=ffffff&fg=000&s=0 "PQ=\sqrt { \left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( -3-3 \right) ^{ 2 }\left[ 4-\left( -4 \right) \right] ^{ 2 } }")
(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is
![PQ=\sqrt { \left( -2-2 \right) ^{ 2 }+\left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( 3-3 \right) ^{ 2 } }](http://s0.wp.com/latex.php?latex=PQ%3D%5Csqrt+%7B+%5Cleft%28+-2-2+%5Cright%29+%5E%7B+2+%7D%2B%5Cleft%5B+1-%5Cleft%28+-1+%5Cright%29+%5Cright%5D+%5E%7B+2+%7D%2B%5Cleft%28+3-3+%5Cright%29+%5E%7B+2+%7D+%7D+&bg=ffffff&fg=000&s=0 "PQ=\sqrt { \left( -2-2 \right) ^{ 2 }+\left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( 3-3 \right) ^{ 2 } }")
Question 2.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.
Now AC = AB + BC
Thus, points A, B and C are collinear.
Question 3.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.
Solution:
(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then
Now, AB = BC
Thus, ABC is an isosceles triangle.
(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then
Now, AC2 = AB2 + BC2
Thus, ABC is a right angled triangle.
(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then
Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.
Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).
Question 5.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.
Solution:
Let P(x, y, z) be any point.
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