NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry Ex 11.2
Ex 11.2 Class 11 Maths 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:
Ex 11.2 Class 11 Maths Question 2:
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Hence, the points P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1) are collinear.
Ex 11.2 Class 11 Maths 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:
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.
Ex 11.2 Class 11 Maths 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 P(x, y, z) be the point that is equidistant from the points A(1, 2, 3) and B(3, 2, -1).
Therefore, PA = PB
Or, PA2 = PB2
(x – 1)2 + (y – 2)2 + (z – 3)2 = (x – 3)2 + (y – 2)2 + (z + 1)2
x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1
-2x – 4y – 6z + 14 = -6x – 4y + 2z + 14
4x – 8z = 0
x – 2z = 0
Thus, the required equation is x – 2z = 0.
Ex 11.2 Class 11 Maths 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:
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