Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Three Dimensional Coordinates Solutions Exercise 5(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Three Dimensional Coordinates Solutions Exercise 5(b)

I.

Question 1.
Find the ratio in which the XZ-plane divides the line joining A(-2, 3, 4) and B(l, 2,3).
Solution:
Ratio in which XZ plane divides
AB = -y1 : y2
= -3 : 2

Question 2.
Find the co-ordinates of the vertex ‘C’ of ∆ABC if its centroid is the origin and the vertices A, B are (1, 1, 1) and (-2, 4, 1) respectively.
Solution:
A(1, 1, 1) B(-2, 4, 1) and (x, y, z) are the vertices of ∆ABC.
G is the centroid of ∆ABC
Co-ordinates of G are
($$\frac{1-2+x}{3}$$, $$\frac{1+4+y}{3}$$, $$\frac{1+1+z}{3}$$) = (0, 0, 0)
$$\frac{x-1}{3}$$ = 0, $$\frac{y+5}{3}$$ = 0, $$\frac{z+2}{3}$$ = 0
x – 1 = 0, y +5 =0, z + 2 = 0
x = 1, y = – 5, z = -2
∴ Co-ordinates of c are (1, -5, -2)

Question 3.
If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of a tetrahedron, find the fourth vertex.
Solution:
A(3, 2, -1), B(4, 1, 1), C(6, 2, 5), D(x, y, z) are the vertices of the tetrahedron.
Co-ordinates of the centroid G are

Co-ordinates of D are (3, 3, 3)

Question 4.
Find the distance between the mid point of the line segment $$\overline{\mathrm{A B}}$$ and the point (3, -1, 2) where A = (6, 3, -4) and B = (-2, -1, 2).
Solution:
A(6, 3, – 4) B(-2, -1,2) are the given points.
Q is the midpoint of AB.
Co-ordinates of Q are ($$\frac{6-2}{2}$$, $$\frac{3-1}{2}$$, $$\frac{-4+2}{2}$$) = (2, 1, -1)
Co-ordinates of P are (3, -1, 2)

II.

Question 1.
Show that the points (5, 4,2) (6,2, -1) and (8, -2, -7) are collinear.
Solution:
A(5, 4, 2), B(6, 2, -1) c(8, -2, -7) are the given points.
AB = $$\sqrt{(5-6)^{2}+(4-2)^{2}+(2+1)^{2}}$$

∴ A, B, C are collinear.

Question 2.
Show that the points A(3, 2, -4), B(5, 4, -6) and C(9, 8, -10) are collinear and find the ratio in which B divides $$\overline{\mathrm{A C}}$$.
Solution:
A(3, 2, -4), B(5, 4, -6) and C(9, 8, -10) are the given points.

A, B, C are collinear
Ratio in which B divides AC = AB : BC
= 2√3 : 4√3 = 1 : 2

III.

Question 1.
If A(4, 8, 12) B(2, 4, 6) C(3, 5, 4) and D(5, 8, 5) are four points, show that the line $$\stackrel{\leftrightarrow}{A B}$$ and $$\stackrel{\leftrightarrow}{C D}$$ intersect.
Solution:
A(4, 8, 12), B(2, 4, 6) C(3, 5, 4) and D(5, 8, 5) are the given points
Co-ordinates of the point dividing AB in the ratio λ : 1 is
$$\left[\frac{2 \lambda+4}{\lambda+1} \cdot \frac{4 \lambda+8}{\lambda+1} \cdot \frac{6 \lambda+12}{\lambda+1}\right]$$ ……….. (1)
Co-ordinates of the point dividing CD in the ratio p : 1 is
$$\left[\frac{5 \mu+3}{\mu+1}, \frac{8 \mu+5}{\mu+1}, \frac{5 \mu+4}{\mu+1}\right]$$ ……….. (2)
If the given lines intersects these two points must coincide.
$$\frac{2 \lambda+4}{\lambda+1}=\frac{5 \mu+3}{\mu+1}$$
(2λ + 4) (µ + 1) = (5µ + 3) (λ + 1)
2λµ + 2λ + 4µ + 4 = 5λµ + 5µ + 3λ + 3
3λ + λ + µ – 1 = 0
λ(3µ+1) = -(µ – 1)
λ = –$$\frac{(\mu-1)}{3 \mu+1}$$
$$\frac{4 \lambda+8}{\lambda+1}=\frac{8 \mu+5}{\mu+1}$$
(4λ + 8) (µ + 1) = (8µ + 5) (λ + 1)
4λµ + 4λ + 8µ + 8 = 8λµ + 8µ + 5λ + 5
4Aµ + λ – 3 = 0
(4µ + 1) λ = 3
$$-\frac{(4 \mu+1)(\mu-1)}{3 \mu+1}=3$$
4µ2 – 4µ + µ – 1 = -9µ – 3
4µ2 4- 6µ + 2 = 0
2µ2 + 3µ + 1=0
(2µ + 1) (µ – 1) = 0
µ = – $$\frac{1}{2}$$ or – 1
µ = -1 is not possible
µ = – $$\frac{1}{2}$$

Co-ordinates of (2) are

Co-ordinates of (1) are
$$\frac{-6+4}{-3+1}, \frac{-12+8}{-3+1}, \frac{-18+12}{-3+1}$$ = (1, 2, 3)
Since these two points coincide the given lines intersect.

Question 2.
Find the point of instersection of the lines $$\stackrel{\leftrightarrow}{A B}$$ and $$\stackrel{\leftrightarrow}{C D}$$ where A = (7, -6, 1) B = (17, -18, -3), C = (1, 4, -5) and D = (3, -4, 11)

Solution:
A(7, -6, 1), B(17, -18, -3), C(1, 4, -5) and D(3, -4, 11) are the given points.
Co-ordinates of the point dividing AB in the ratio λ: 1 are

Co-ordinates of the point dividing CD in the ratio µ : 1 are

(17λ + 7) (µ + 1) = (3µ + 1) (λ + 1)
17λµ + 17λ + 7µ + 7 = 3λµ + 3µ + λ + 1
14λµ + 16λ + 4µ + 6 = 0.

-18λµ – 6µ – 18λ – 6
= -4λµ + 4λ – 4µ + 4
14λµ + 22λ + 2µ + 10 = 0
14λµ + 16λ + 4µ + 6 = 0 ………. (1)
14λµ + 22λ + 2µ + 10 = 0……….(2)
Subtraction – 6λ + 2µ – 4 = 0
2µ = 6λ + 4
µ = 3λ + 2
Substituting in (3)
14λ(3λ + 2) + 16λ + 4(3λ + 2) + 6 = 0
42λ² + 28λ + 16λ + 12λ+ 8 + 6 = 0
42λ² + 56λ + 14 = 0
3λ² + 4λ + 1 = 0
(λ + 1) (3λ + 1) = 0
λ = -1 or λ = –$$\frac{1}{3}$$
λ = -1 is not possible
λ = –$$\frac{1}{3}$$

µ = 3λ + 2 = -1 + 2 = 1
µ = 1 ⇒
$$\frac{3+1}{1+1}, \frac{-4+4}{1+1}, \frac{11-5}{1+1}$$ = (2, 0, 3)
∴ These two points coincide
⇒ The given lines AB and CD intersect
Point of intersection is (2, 0, 3)

Question 3.
A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3) are vertices of a triangle. $$\overline{\mathrm{A D}}$$, the bisector of ∠BAC meets $$\overline{\mathrm{B C}}$$ at D. Find the co-ordinates of D.
Solution:
A(3, 2, 0),B(5, 3, 2) C(-9, 6, -3) are the vertices of ∆ABC

AB is the bisector of ∠DAC
‘D’ divides BC in the ratio 3 : 13
Co-ordinates of D are

Question 4.
Show that the points 0(0,0,0), A(2, -3, 3), B(-2, 3, -3) are collinear. Find the ratio in which each point divides the segment joining the other two.
Solution:
0(0, 0, 0), A(2, -3, 3), B(-2, 3, -3) are the given points.

∴ O, A, B are collinear.

Ratio in which ‘O’ divides AB
= OA : OB = √22 : √22 = 1 : 1
Ratio is which A divides OB
= OA:AB = -√22 : √22 = -1 : 2
Ratio in which B divides OA
= AB : BO = -2√22 : √22 =-2 : 1
A and B divide externally.