In Exercise 28.3 of Chapter 28, we shall discuss some problems related to section formulae. From the exam point of view, solutions are formulated by our subject expert team and contain explanations in simple language, which is very helpful for students to understand the problems. Students can improve their problem-solving ability by practising the problems provided in the RD Sharma Class 11 Maths Solutions which is very beneficial during their exam preparation. Students can download the PDF from the available links given below.
RD Sharma Solutions for Class 11 Maths Exercise 28.3 Chapter 28 – Introduction to Three-Dimensional Coordinate Geometry
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ExERCISE 28.3 PAGE NO: 28.19
1. The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.
Solution:
Given:
The vertices of the triangle are A (5, 4, 6), B (1, -1, 3) and C (4, 3, 2).
By using the formulas, let us find the coordinates of D and the length of AD
The distance between any two points (a, b, c) and (m, n, o) is given by,
The section formula is given as
AB : AC = 5:3
BD: DC = 5:3
So, m = 5 and n = 3
B(1, -1, 3) and C(4, 3, 2)
Coordinates of D using section formula:
2. A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10). Find the coordinates.
Solution:
Given:
Points A (2, -3, 4) and B (8, 0, 10)
By using the section formula,
Let Point C(x, y, 8), and C divides AB in ratio k: 1
So, m = k and n = 1
A(2, -3, 4) and B(8, 0, 10)
Coordinates of C are:
On comparing, we get,
[10k + 4] / [k + 1] = 810k + 4 = 8(k + 1)
10k + 4 = 8k + 8
10k – 8k = 8 – 4
2k = 4
k = 4/2
= 2
Here C divides AB in a ratio 2:1
x = [8k + 2] / [k + 1]
= [8(2) + 2] / [2 + 1]
= [16 + 2] / [3]
= 18/3
= 6
y = -3 / [k + 1]
= -3 / [2 + 1]
= -3 / 3
= -1
∴ The Coordinates of C are (6, -1, 8).
3. Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.
Solution:
Given:
Points A (2, 3, 4), B (-1, 2, -3) and C (-4, 1, -10)
By using the section formula,
Let C divides AB in ratio k: 1
Three points are collinear if the value of k is the same for x, y and z coordinates.
So, m = k and n = 1
A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)
Coordinates of C are:
On comparing, we get,
[-k + 2] / [k + 1] = -4-k + 2 = -4(k + 1)
-k + 2 = -4k – 4
4k – k = – 2 – 4
3k = -6
k = -6/3
= -2
[2k + 3] / [k + 1] = 12k + 3 = k + 1
2k – k = 1 – 3
k = – 2
[-3k + 4] / [k + 1] = -10-3k + 4 = -10(k + 1)
-3k + 4 = -10k – 10
-3k + 10k = -10 – 4
7k = -14
k = -14/7
= -2
The value of k is the same in all three cases.
So, A, B and C are collinear [as k = -2]
∴We can say that C divides AB externally in a ratio 2:1
4. Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.
Solution:
Given:
The points (2, 4, 5) and (3, 5, 4)
By using the section formula,
We know the X coordinate is always 0 on yz-plane
So, let Point C(0, y, z), and let C divide AB in ratio k: 1
Then, m = k and n = 1
A(2, 4, 5) and B(3, 5, 4)
The coordinates of C are:
On comparing, we get,
[3k + 2] / [k + 1] = 03k + 2 = 0(k + 1)
3k + 2 = 0
3k = – 2
k = -2/3
∴We can say that C divides AB externally in a ratio 2: 3
5. Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.
Solution:
Given:
The points(2, -1, 3) and (-1, 2, 1)
By using the section formula,
Let C(x, y, z) be any point on the given plane, and C divides AB in ratio k: 1
Then, m = k and n = 1
A(2, -1, 3) and B(-1, 2, 1)
Coordinates of C are:
On comparing, we get,
[-k + 2] / [k + 1] = x [2k – 1] / [k + 1] = y [-k + 3] / [k + 1] = zWe know that x + y + z = 5
5(k + 1) = 4
5k + 5 = 4
5k = 4 – 5
5k = – 1
k = -1/5
∴We can say that the plane divides AB externally in the ratio 1:5
6. If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.
Solution:
Given:
Points A (3, 2, -4), B (9, 8, -10) and C (5, 4, -6)
By using the section formula,
Let C divides AB in ratio k: 1
Three points are collinear if the value of k is the same for x, y and z coordinates.
Then, m = k and n = 1
A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)
Coordinates of C are:
On comparing, we get,
[9k + 3] / [k + 1] = 59k + 3 = 5(k + 1)
9k + 3 = 5k + 5
9k – 5k = 5 – 3
4k = 2
k = 2/4
= ½
[8k + 2] / [k + 1] = 48k + 2 = 4(k + 1)
8k + 2 = 4k + 4
8k – 4k = 4 – 2
4k = 2
k = 2/4
= ½
[-10k – 4] / [k + 1] = -6-10k – 4 = -6(k + 1)
-10k – 4 = -6k – 6
-10k + 6k = 4 – 6
-4k = -2
k = -2/-4
= ½
The value of k is the same in all three cases.
So, A, B and C are collinear [as, k = ½]
∴We can say that C divides AB externally in a ratio 1:2
7. The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.
Solution:
Given:
The mid-points of the sides of a triangle ABC are given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).
By using the section formula,
We know the mid-point divides sides in the ratio of 1:1.
The coordinates of C are given by,
P(-2, 3, 5) is mid-point of A(x1, y1, z1) and B(x2, y2, z2)
Then,
Q(4, -1, 7) is mid-point of B(x2, y2, z2) and C(x3, y3, z3)
Then,
R(6, 5, 3) is mid-point of A(x1, y1, z1) and C(x3, y3, z3)
Then,
Now solving for ‘x’ terms
x1 + x2 = -4……………………(4)
x2 + x3 = 8………………………(5)
x1 + x3 = 12……………………(6)
By adding equations (4), (5), (6)
x1 + x2 + x2 + x3 + x1 + x3 = 8 + 12 – 4
2x1 + 2x2 + 2x3 = 16
2(x1 + x2 + x3) = 16
x1 + x2 + x3 = 8………………………(7)
Now, subtract equations (4), (5) and (6) from equation (7) separately:
x1 + x2 + x3 – x1 – x2 = 8 – (-4)
x3 = 12
x1 + x2 + x3 – x2 – x3 = 8 – 8
x1 = 0
x1 + x2 + x3 – x1 – x3 = 8 – 12
x2 = -4
Now solving for ‘y’ terms
y1 + y2 = 6……………………(8)
y2 + y3 = -2……………………(9)
y1 + y3 = 10……………………(10)
By adding equations (8), (9) and (10) we get,
y1 + y2 + y2 + y3 + y1 + y3 = 10 + 6 – 2
2y1 + 2y2 + 2y3 = 14
2(y1 + y2 + y3) = 14
y1 + y2 + y3 = 7………………………(11)
Now, subtract equations (8), (9) and (10) from equation (11) separately:
y1 + y2 + y3 – y1 – y2 = 7 – 6
y3 = 1
y1 + y2 + y3 – y2 – y3 = 7 – (-2)
y1 = 9
y1 + y2 + y3 – y1 – y3 = 7 – 10
y2 = -3
Now solving for ‘z’ terms
z1 + z2 = 10……………………(12)
z2 + z3 = 14……………………(13)
z1 + z3 = 6……………………(14)
By adding equations (12), (13) and (14) we get,
z1 + z2 + z2 + z3 + z1 + z3 = 6 + 14 + 10
2z1 + 2z2 + 2z3 = 30
2(z1 + z2 + z3) = 30
z1 + z2 + z3 = 15………………………(15)
Now, subtract equations (8), (9) and (10) from equation (11) separately:
z1 + z2 + z3 – z1 – z2 = 15 – 10
z3 = 5
z1 + z2 + z3 – z2 – z3 = 15 – 14
z1 = 1
z1 + z2 + z3 – z1 – z3 = 15 – 6
z2 = 9
∴The vertices of sides of a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).
8. A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.
Solution:
Given:
The vertices of a triangle are A (1, 2, 3), B (0, 4, 1), C (-1, -1, -3)
By using the distance formula,
So, AB/AC = 3/7
AB: AC = 3:7
BD: DC = 3:7
Then, m = 3 and n = 7
B(0, 4, 1) and C(-1, -1, -3)
Coordinates of D by using the section formula are given as
∴The coordinates of D are (-3/10, 5/2, -1/5).
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