CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths
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2 2 2
DA 1 2 23 14
925 9 43
Here, AB = CD = 6, BC = AD = 43
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.
Question 4:
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -
1).
Solution 4:
Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).
Accordingly, PA = PB
2 2
PA
PB
x 1 y 2 z 3 x 3 y 2 z
1
2 2 2 2 2 2
⇒ x 2 – 2x + 1 + y 2 – 4y + 4 + z 2 – 6z + 9 = x 2 – 6x + 9 + y 2 – 4y + 4 + z 2 + 2z + 1
⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.
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 5:
Let the coordinates of P be (x, y, z).
The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.
It is given that PA + PB = 10.
2 2 2 2 2 2
x 4 y z x 4 y z 10
4 10 4
2 2 2 2 2 2
x y z x y z
On squaring both sides, we obtain
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