OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

Question 1.
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Solution:
Let P be any point on y-axis and its coordinates are (0, y, 0) and given points are A(3, 1, 2) and B (5, 5, 2).
According to given condition; we have

On squaring both sides; we have
9 + (1 – y)² + 4 = 25 + (5 – y)² + 4
⇒ 14 + y² – 2y = 54 + y² – 10y
⇒ 8y = 40
⇒ y = 5
Thus required point on y-axis be P(0, 5, 0).

Question 2.
Show that the points (a, b, c),(b, c, a) and (c, a, b) are the vertices of an equilateral trinagle.
Solution:
Let (a, b, c); (b, c, a) and (c, a, b) are the given vertices A, B and C of △ABC.

Thus, △ABC be an equilateral triangle.

Question 3.
Find out whether the points (0, 7, -10), (1, 6, 6) and (4, 9, -6) are the vertices of a right angled triangle.
Solution:
Let A(0,7,-10) ; B(1, 6, 6) and C(4, 9, -6) are the vertices of △ABC respectively.
AB² = (1 – 0)² + (6 – 7)² + (6 + 10)² = 1 + 1 +256 = 258
BC² = (4 – 1)² + (9 – 6)² + (- 6 – 6)² = 9 + 9 + 144 = 162
CA² = (4 – 0)² + (9 – 7)² + (- 6 + 10)² = 16 + 4 + 16 = 36
Thus,
AB² ≠ BC² + AC²
BC² ≠ AB² + AC²
AC² ≠ AB² + BC²
Thus, pythagoras theorem does not holds good. ∴ △ABC is not a right angled Δ.

Question 4.
Show that the points A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) are collinear.
Solution:
Given points are A(-2, 3, 5); B(1, 2, 3) and C(7, 0, -1)
Here AB = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}\) = \(\sqrt{9+1+4}\) = \(\sqrt{14}\)
BC = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\) = \(\sqrt{36+4+16}\) = \(\sqrt{56}\)
and CA = \(\sqrt{(-2-7)^2+(3-0)^2+(5+1)^2}\) = \(\sqrt{81+9+36}\) = \(\sqrt{126}\) = \(3 \sqrt{14}\)
Thus AB + BC + CA
Therefore all the given points A, B and C lies on same straight line and hence given points are collinear.

Question 5.
Find the lengths of the medians of the tri- angle A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0)
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.
Thus coordinates of D, E and F using mid-point formula be given by D(3, 2, 0); E(3, 0, 3) and F(0, 2, 3).

Question 6.
Using section formula, show that the points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) are collinear.
Solution:
Let A, B and C are the given points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) respectively.
Thus the coordinates of the point divides the straight line joining the points B and C in the ratio k : 1 internally are

∴ 2 = –\(\frac{1}{k+1}\) ⇒ k + 1 = –\(\frac{1}{2}\) ⇒ k = –\(\frac{3}{2}\)
\(\frac{\frac{k}{3}+2}{k+1}\) = – 3 ⇒ \(\frac{k}{3}\) + 2 = – 3k – 3
⇒ \(\frac{10k}{3}\) = -5 ⇒ k = –\(\frac{3}{2}\)
and \(\frac{2 k+1}{k+1}\) ⇒ 2k + 1 = 4k + 4
⇒ 2k = – 3 ⇒ k = –\(\frac{3}{2}\)
Thus from all equations, we get same value of k.
Hence, the point A lies on the straight line joining B and C. Thus, points A, B and C are collinear.

Question 7.
Find the ratio in which the yz-plane divides the line segment formed by joining the point (-2, 4, 7) and (3, -5, 8).
Solution:
Let the point P divides the line segment joining A(-2, 4, 7) and B(3, -5, 8) in the ratio k : 1.

Clearly the line segment AB is divided by yz-plane so x-coordinate of point P is 0.
∴ \(\frac{3 k-2}{k+1}\) = 0 ⇒ k = \(\frac{2}{3}\)
Thus, required ratio be k : 1 i.e. 2 : 3