OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Students often turn to ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 1.
Find the coordinates of the points which divide the join of the points (2, -1, 3) and (4, 3, 1) in the ratio 3 : 4 internally.
Solution:
Let R be the point which divides the join of points A(2, -1, 3) and B(4, 3, 1) in the ratio 3 : 4 internally.

Question 2.
Find the coordinates of the points which divide the line joining the points (2, -4, 3), (-4, 5, -6) in the ratio
(i) 1 : -4
(ii) 2 : 1
Solution:
(i) Let the point P divides the line segment AB in ratio 1 :-4
Then coordinates of P are

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

Since the line joining AB is divided by yz – plane
i.e. x-coordinates of point P is 0 .
\(\frac{3 k+2}{k+1}\) = 0
⇒ k = –\(\frac{2}{3}\)
Thus required ratio be k : 1 i.e. –\(\frac{2}{3}\) : 1 i.e. – 2 : 3.

Question 4.
The three points A(0, 0, 0), B(2, -3, 3), C(-2, 3, -3) are collinear. Find in what ratio each point divides the segment joining the other two.
Solution:
Let the point B (2, -3, 3) divides the line segment AC in the ratio k : 1 internally.
Then by section formula, we have
The coordinates of B are

∴ \( \frac{-2 k}{k+1}\) = 2
⇒ – 2k = 2k + 2
⇒ 4k = – 2
⇒ k = \(\frac{-1}{2}\)
and \(\frac{3 k}{k+1}\) = – 3
⇒ 3k = -3 k – 3
⇒ 6 k = – 3
⇒ k =-\(\frac{1}{2}\)
and \(\frac{-3 k}{k+1}\) = 3
⇒ – 3k = 3k + 3
⇒ k = –\(\frac{1}{2}\)
Thus the required ratio be k : 1 i.e. – 1 : 2.
Let the point C(-2, 3, -3) divides AB in the ratio λ : 1.
Then coordinates of C are

\(\frac{2 \lambda}{\lambda+1}\) = – 2 ⇒ 2λ = – 2λ – 2 ⇒ λ = –\(\frac { 1 }{ 2 }\)
\(\frac{-3 \lambda}{\lambda+1}\) = 3 ⇒ -3λ = 3λ + 3 ⇒ λ = –\(\frac { 1 }{ 2 }\)
and \(\frac{3 \lambda}{\lambda+1}\) = – 3 ⇒ 6λ = – 3
Thus required ratio be λ : 1 i.e. -1 : 2. Let the point A(0, 0, 0) divides line segment BC in the ratio p : 1.

Question 5.
Find the coordinates of the points which trisect AB given that A(2, 1, -3) and B (5, -8, 3).
Solution:
Let P and Q be the point of trisection of line segment AB.

Thus point P divides the line segment AB in the ratio 1 : 2.
Then coordinates of P are
\(\left(\frac{5+4}{1+2}, \frac{-8+2}{1+2}, \frac{3-6}{1+2}\right)\) i.e. P(3, -2, -1)
Also, point Q divides the line segment AB in the ratio 2 : 1.
Then coordinates of $\mathrm{Q}$ are
\(\left(\frac{10+2}{2+1}, \frac{-16+1}{2+1}, \frac{6-3}{2+1}\right)\) i.e. Q(4, -5, 1).

Question 6.
Find the coordinates of the point which is three-fifths of the way from (3, 4, 5) to (-2, -1, 0).
Solution:
Let the given points are A(3, 4, 5) and E(-2, -1, 0) and let point P is at \(\frac{3}{5}\)th of the way from A.
∴ P is at a \(\frac{2}{5}\)th of the way from B i.e. p divides line segment AB in the ratio 3 : 2.
Then by section formula, we have

Thus, the coordinates of point P are (0, 1, 2).

Question 7.
Show that the point (1, -1, 2) is common to the lines which join (6, -7), 0) to (16, -19, -4) and (0, 3, -6) to (2, -5, 10).
Solution:
Any point on line segment AB be

If AB and CD have a common point. Then P and Q coincide for some values of k and k’.

Thus eqns. (1), (2) and (3) are satisfied or consistent for k = \(\frac{-1}{3}\) and k’ = 1
putting k = \(\frac{-1}{3}\) in coordinates of P
we get, the required point be (1, -1, 2). Hence, the point P(1, -1, 2) is common to lines which join A(6, – 7, 0)
and B(16, -19, -4) and C(0, 3, -6) and D(2, -5, 10).

Question 8.
Find the lengths of the medians of the triangle whose vertices are A(2, -3, 1), B (-6, 5, 3), C (8, 7, – 7).
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.

Question 9.
Find the point of intersection of the medians of the triangle with vertices (-1, -3, -4), (4, -2, -7), (2, 3, -8).
Solution:
Let the vertices of △ABC are A(-1, -3, -4); B(4, -2, -7) and C(2, 3, -8)
We know that the point of intersection of all medians of a triangle is called centroid of triangle.

Question 10.
Find the ratio in which the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane 2x + 2y – 2z = 1. Also, find the coordinates of the point of division.
Solution:
Let the point R divides the line segment PQ in the ratio k : 1 internally.

Then by section formula, we have coordinates of R are
\(\left(\frac{3 k+2}{k+1}, \frac{4 k+1}{k+1}, \frac{3 k+5}{k+1}\right)\)
Clearly it is given that, line segment PQ is divided by the plane
2x + 2y – 2z = 1 …(1)
Thus the point R lies on eqn. (1); we have
\(2\left(\frac{3 k+2}{k+1}\right)+2\left(\frac{4 k+1}{k+1}\right)-2\left(\frac{3 k+5}{k+1}\right)\) = 1
⇒ 6k + 4 + 8k + 2 – 6k – 10 = k + 1
⇒ 7k = 5 ⇒ k = \(\frac{5}{7}\)
Thus the required ratio be k : 1
i.e. \(\frac{5}{7}\) : 1 i.e. 5 : 7
Thus, required point of division be

Question 11.
The mid-points of the sides of astriangle are (1, 5, -1),(0, 4, -2) and (2, 3, 4). Find its vertices.
Solution:
Let the vertices of △ABC are A (x₁, y₁, z₁);
B (x₂, y₂, z₂) and
C (x₃, y₃, z₃).
It is given that D (1, 5, -1) ; E(0, 4, – 2) and F(2, 3, 4) are the mid-points of sides BC, CA and AB of △ABC.
Now D be the mid-point of BC.

On adding eqn. (1), (4) and (7); we have
x₁ + x₂ + x₃ = 3 …(10)
From eqn. (1) and eqn. (10); x₁ = 1
From (4) and (10); x₂ = 3
and From eqn. (7) and (10); x₃ = -1
On adding eqn. (2), (5) and (8); we have
y₁ + y₂ + y₃ = 12 …(11)
From (2) and (11); y₁ = 2
From (5) and (11); y₂ = 4
From (8) and (11); y₃ = 6
On adding eqn. (3), (6) and (9); we have
z₁ + z₂ + z₃ = 1 …(12)
From (3) and (12); z₁ = 3
From (6) and (12); z₂ = 5
From (9) and (12); z₃ = -7
Thus the required vertices of △ABC are (1, 2, 3) ;(3, 4, 5) and (-1, 6, -7).

Question 12.
Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex D.
Solution:
Given vertices of parallelogram ABCD are A(3, -1, 2); B(1, 2, -4); C(-1, 1, 2) and let the coordinates of fourth vertex D are (α, β ,γ).
Mid-point of AC = \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)\)
and Mid-point of BD = \(\left(\frac{1+\alpha}{2}, \frac{2+\beta}{2}, \frac{-4+\gamma}{2}\right)\)
Since ABCD be a parallelogram.
∴ diagonals of || gm ABCD bisect each other.
Thus, mid-point of AC = mid-point of BD

Thus the coordinates of fourth vertex D are $(1,-2,8)$.

Question 13.
What is the locus of a point for which
(i) x = 0
(ii) y = 0
(iii) z = 0
(iv) x =a
(v) y = b
(vi) z = c?
Solution:
(i) The point for which x = 0 is of the form (0, y, z) and lies in yoz plane. Thus required locus of point be yz-plane.

(ii) The point for which y = 0 is of the form (x, 0, z) and lies in the xoz plane.
Thus, required locus of point be xz-plane.

(iii) The point for which z = 0 is of the form (x, y, 0) and lies in xoy plane. Thus required locus of a point be xy-plane.

(iv) Since x = a be the plane parallel to x = 0 i.e. yz plane. Thus locus of a point for which x = a be a plane parallel to yz plane at a distance of a units from it.

(v) Since y = b be the plane parallel to y = 0 i.e. xz-plane. Thus, locus of a point for which y = b be a plane parallel to xz plane at a distance b units from it.

(vi) Since z = c be a plane parallel to plane z = 0 i.e. xy plane. Thus locus of a point for which z = c be a plane || to xy plane at a distance c units from it.

Question 14.
What is the locus of a point for which
(i) x = 0, y = 0
(ii) y = 0, z = 0
(iii) z = 0, x = 0
(iv) x = a, y = b
(v) y = b, z = c
(vi) z = c, x = a?
Solution:
(i) We know that on z-axis, x = 0 = y Thus required locus be z-axis.
(ii) We know that on x-axis, we have y = 0 = z Thus required locus be x-axis.
(iii) We know that on y-axis, we have x = 0 = z Thus, required locus be y-axis.
(iv) x = a be the line || to y-axis and y = b be the line || to x-axis
Thus locus of a point for which x = a, y = b is the line of intersection of given planes x = a and y = b
(v) Clearly the locus of a point for which y = b, z = c is the line of intersection of given planes y = b and z = c
(vi) Given planes are z = c and x = a
Required locus is the line of intersection of planes z = c and x = a.