RD Chapter 28 Introduction to 3D coordinate geometry Ex 28.3 Solutions
Question - 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.
Answer - 1 : -
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 thelength 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:
Question - 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.
Answer - 2 : -
Given:
The 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] = 8
10k + 4 = 8(k + 1)
10k + 4 = 8k + 8
10k – 8k = 8 – 4
2k = 4
k = 4/2
= 2
Here C divides AB in 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).
Question - 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.
Answer - 3 : -
Given:
The 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] = 1
2k + 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 ratio 2:1
Question - 4 : - Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.
Answer - 4 : -
Given:
The points (2, 4, 5) and (3, 5, 4)
By using the section formula,
We know 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] = 0
3k + 2 = 0(k + 1)
3k + 2 = 0
3k = – 2
k = -2/3
∴We can say that, C divides AB externally in ratio 2: 3
Question - 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.
Answer - 5 : -
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] = z
We know that x + y + z = 5
5(k + 1) = 4
5k + 5 = 4
5k = 4 – 5
5k = – 1
k = -1/5
∴We can saythat, the plane divides AB externally in the ratio 1:5
Answer - 6 : -
Given:
The points A (3, 2,-4), B (9, 8, -10) and C (5, 4, -6)
By using the sectionformula,
Let C divides AB inratio k: 1
Three points arecollinear 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 ofC are:
On comparing we get,
[9k +3] / [k + 1] = 5
9k + 3 = 5(k + 1)
9k + 3 = 5k + 5
9k – 5k = 5 – 3
4k = 2
k = 2/4
= ½
[8k +2] / [k + 1] = 4
8k + 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 thesame in all three cases.
So, A, B and Care collinear [as, k = ½]
∴We can say that, Cdivides AB externally in ratio 1:2
Question - 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.
Answer - 7 : -
Given:
The mid-points of thesides of a triangle ABC is given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).
By using the sectionformula,
We know the mid-pointdivides side in the ratio of 1:1.
The coordinates of Cis given by,
P(-2, 3, 5) ismid-point of A(x1, y1, z1) and B(x2,y2, z2)
Then,
Q(4, -1, 7) ismid-point of B(x2, y2, z2) and C(x3,y3, z3)
Then,
R(6, 5, 3) ismid-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 equation(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 equation(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 equation(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 equation(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 equation(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 equation(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
∴Thevertices of sidesof a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).
Question - 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.
Answer - 8 : -
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 byusing section formula is given as
∴The coordinates of Dare (-3/10, 5/2, -1/5).