Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(b)

I.

Question 1.
Find the direction ratios of the line joining the points (3, 4, 0) and (4, 4, 4).
Solution:
A(3, 4, 0) and B(4, 4, 4) are the given points.
d.rs of AB are (x2 – x1, y2 – y1, z2 – z1)
(4-3, 4-4, 4-0) i.e.,(1, 0, 4)

Question 2.
The direction ratios of a line are (-6, 2, 3). Find its direction cosines.
Solution:
D.Rs of the line are -6, 2, 3
Dividing with \(\sqrt{36+4+9}\) = 7
Direction cosines of the line are – \(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\)

Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b)

Question 3.
Find the cosine of the angle between the lines whose direction cosines are (\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)) and (\(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\), 0)
Solution:
cos θ = l1 + l2 + m1 m2 + n1 n2
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 1

Question 4.
Find the angle between the lines whose direction ratios are (1, 1, 2) (√3, -√3, 0).
Solution:
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 2

Question 5.
Show that the lines with direction cosines (\(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\)) and (\(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\)) are perpendicular to each other.
Solution:
If two lines, are perpendicular, then
l1l2 + m1m2 + n1n2 = 0
l1l2 + m1m2 + n1n2
\(\frac{12}{13} \cdot \frac{4}{13}-\frac{3}{13} \cdot \frac{12}{13}-\frac{4}{13} \cdot \frac{3}{13}\)
∴ The given lines are perpendicular.

Question 6.
O is the origin, P(2, 3, 4) and Q(1, k, 1) are points such that \(\overline{\mathrm{O P}}\) ⊥ \(\overline{\mathrm{O Q}}\). Find k.
Solution:
d.rs of OP and 2, 3,4
d.rs of OQ are 1, k, 1
OP and OQ are perpendicular
⇒ a1a2 + b1b2 + c1c2 = 0
2 + 3k + 4 = 0
3k = -6
k = -2.

II.

Question 1.
If the direction ratios of a line are (3, 4, 0), find its direction cosines are also the angles made with the co-ordinate axes.
Solution:
Direction ratios of the line are (3, 4, 0)
Dividing with \(\sqrt{9+16+0}\) = 5
D.C’s of the line are (\(\frac{3}{5}, \frac{4}{5}\), 0)
If αc, βc² are the angles made By the line with the co-ordinate axes, then
cos α = \(\frac{3}{5}\) cos β = \(\frac{4}{5}\) cos γ = 0
α = cos-1\(\frac{3}{5}\), β = cos-1\(\frac{4}{5}\), γ = \(\frac{\pi}{2}\)
Angles made with co-ordinate axes are
cos-1\(\frac{3}{5}\), cos-1\(\frac{4}{5}\), γ = \(\frac{\pi}{2}\)

Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b)

Question 2.
Show that the line through the points (1, -1,2) (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
A(1, -1 2) B(3, 4, -2) C(0,3,2) and D(3, 5,6) are the given points
d.rs of AB are 3 – 1, 4 + 1, -2 -2 i.e., 2, 5, -4
d.rs of CD are 3 – 0, 5 – 3, 6 – 2 i.e., 3, 2 4
a1a2 + b1b2 + c1c2 = 2.3 + 5.2 – 4.4
= 6 + 10 – 16 = 0
AB and CD are perpendicular.

Question 3.
Find the angle between \(\overline{\mathrm{D C}}\) and \(\overline{\mathrm{A B}}\) where A = (3, 4, 5), B =(4, 6, 3) C = (-1, 2, 4) are D(l, 0, 5).
Solution:
A(3, 4, 5), B(4, 6, 3), C(-1, 2, 4), D(1, 0, 5) are the given points.
d.rs of AB are 4 – 3, 6 – 4, 3 – 5 i.e., 1, 2, -2
d.rs of CD are 1 + 1, 0 – 2, 5 – 4 i.e., 2, -2, 1
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 3

Question 4.
Find the direction cosines of a line which is perpendicular to the lines, whose direction ratios are (1, -1, 2) and (2, 1, -1).
Solution:
Let l, m, n be the d.cs of the required line it is perpendicular to the lines with
d.rs (1, -1, 2) and (2, 1, -1)
l – m + 2n = 0
2l + m – n = 0
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 4
d.rs of the line are – 1, 5, 3
Dividing with \(\sqrt{1+25+9} =\sqrt{35}\)
d.cs of the required line are ’
– \(\frac{1}{\sqrt{35}}, \frac{5}{\sqrt{35}}, \frac{3}{\sqrt{35}}\)

Question 5.
Show that the points (2, 3, -4), (1, -2, 3) and (3, 8, -11) are collinear.
Solution:
A(2, 3, -4),B( 1, -2, 3) and C(3, 8, -11) are the given points.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 5
∴ A, B, C are collinear.

Question 6.
Show that the points (4, 7, 8), (2, 3, 4), (-1, -2,1), (1,2,5) are vertices of apare- llelogram.
Solution:
A(4, 7, 8), B(2, 3, 4), C(-1, -2, 1) and D (1, 2, 5) . are the given points.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 6
∴ AB = CD and BC = DA
∴ A, B, C, D are the vertical of parallelogram.

III.

Question 1.
Show that the lines whose direction co – sines are given by l + m + n = 0, 2mn + 3nZ – 5Zm = 0 are perpendicular to each other.
Solution:
Given l + m + n = 0 ………… (1)
2mn + 3nl – 5lm = 0 …………… (2)
From (1), l = -(m + n)
Substituting in (2)
2mn – 3n(m + n) + 5m(m + n) = 0
2mn – 3mn – 3n2 + 5m2 + 5mn = 0
5m² + 4mn – 3n² = 0
\(\frac{m_{1} m_{2}}{n_{1} n_{2}}=-\frac{3}{5} \Rightarrow \frac{m_{1} m_{2}}{-3}=\frac{n_{1} n_{2}}{5}\) ………. (1)
From (1), n = -(l + m).
Substituting is (2)
-2m (l + m) – 3l(l + m) – 5lm = 0
– 2lm – 2m2 – 3l² – 3lm – 5lm = 0
3l² + 10lm + 2m² = 0
\(\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{2}{3} \Rightarrow \frac{l_{1} l_{2}}{2}=\frac{m_{1} m_{2}}{3}\) ……… (2)
From (1) and (2) we get
\(\frac{I_{1} I_{2}}{2}=\frac{m_{1} m_{2}}{3}=\frac{n_{1} n_{2}}{-5}=1\)
l1l2 = 2k, m1m2 = 3k, n1 n2 = -5k
∴ l1l2 + m1m2 + n1n2 = 2k + 3k – 5k = 0
The two lines are perpendicular.

Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b)

Question 2.
Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l² + m² – n² = 0.
Solution:
Given l + m + n = 0
l² + m² – n² = 0
From (1), l = – (m + n)
Substituting in (2)
(m + n)² + m² – n² = 0
m² + n² + 2mn + m² – n² = 0
2m² + 2mn = 0
2m(m + n) = 0
∴ m = 0 or m + n = 0

Case (i) : m = 0, Substituting in (1) l + n = 0
l = -n ⇒ \(\frac{l}{1}\) = \(\frac{n}{-1}\)
D.Rs of the first line l1 are (1, 0, – 1)

Case (ii) :m + n = 0 ⇒ m = -n ⇒ \(\frac{m}{1}\) = \(\frac{n}{-1}\)
Substituting in (1) l = 0
D.Rs of the second line l2 are (0, 1 – 1)
Suppose θ is the angle between the two lines
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 7

Question 3.
If a ray makes angles α, α, γ and δ with the four diagonals of a cube find cos² α + cos² β + cos² γ + cos² δ.
Solution:
Let each side of the cube be of length a. Let one of the vertices of the cube be the origin O and the co-ordinate axes be along the three edges \(\overline{\mathrm{O A}}\), \(\overline{\mathrm{O B}}\) and \(\overline{\mathrm{O C}}\) passing through the origin. The co-ordinates of the vertices of the cube with respect to the frame of reference OABC are as shown in figure. The diagonals of he cubeare \(\overline{\mathrm{O P}}\) \(\overline{\mathrm{C D}}\), \(\overline{\mathrm{A E}}\) and \(\overline{\mathrm{B F}}\) .(a, a, a), (a, a, -a), (-a, a, a) and (a, -a, a) are direction . ratios of these diagonals respectively.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 8

Let the direction cosines of the given ray be (l, m, n). If this fay makes the angles α, β, γ and δ with the four diagonals of the cube, then
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 9
cos² α + cos² β + cos² γ + cos² δ
\(\frac{1}{3}\){|l + m + n|² +|l + m – n|² + |-l + m + n|² + |l – m + n|²}
\(\frac{1}{3}\)[(l + m + n)² + (l + m – n)² + (-l + m + n)² + (l – m + n)²]
\(\frac{1}{3}\)[4(l² + m² + n²)] = \(\frac{4}{3}\) (since l² + m² + n² = 1)

Question 4.
If (l1, m1, n1), (l2, m2, n2) are d.c.s of two intersecting lines, show that d.c.s of two lines, bisecting the angles between them are proportional to l1 ± l2, m1 ± m2 n1 ± n2.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 10
Solution:
OA and OB are the given lines
A and B are points at unit distances from O
Co-ordinates of A are (l1, m1, n1)
Co-ordinates of B are (l2, m2, n2)
P is the mid point of AB
Co-ordinates of P are \(\left(\frac{l_{1}+l_{2}}{2}, \frac{m_{1}+m_{2}}{2}, \frac{n_{1}+n_{2}}{2}\right)\)
∴ OP is the bisector of ∠AOB
D.Rs of OP are l1 + l2, m1+ m2, n1+ n2
Suppose B’ is a point are OB such that OB’ OB = 1
Co-ordinates of B’ are (-l2, -m2, -n2)
Q is the mid point of AB’
Co-ordinates of Q are \(\left(\frac{l_{1}-l_{2}}{2}, \frac{m_{1}-m_{2}}{2}, \frac{n_{1}-n_{2}}{2}\right)\)
OQ is the other bisector of ∠AOB
D.Rs of OQ are l1 – l1, m1 – m2, n1 – n2

Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b)

Question 5.
A (-1, 2, -3), B(5, 0, -6), C(0, 4, -1) are three points. Show that the direction cosines of the bisectors of ∠BAC are propotional to (25, 8, 5) and (-11, 20, 23).
Solution:
A (-1, 2, -3), B (5, 0, -6) and C (0, 4, -1) are the given points.
D.Rs of AB are 5 +1, 0 -2, -6 +3 ie., 6, -2, -3
D.Rs of AB are \(\frac{6}{7}, \frac{-2}{7}, \frac{-3}{7}\)
D.Rs of AC are 0 + 1, 4 – 2, -1 + 3 i.e., 1, 2, 2
D.Rs of AC are \(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\)
D.Rs of one of the bisectors are
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 11
D.Rs of one of the bisectos are (25, 8, 5)
D.Rs of the other bisectors are
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 12
D.Rs of the second bisector are (-11, 20, 23)

Question 6.
If (6,10,10), (1, 0, -5), (6, -10, 0) are vertices of a triangle, find the direction ratios of its sides. Determine whether it is right angled or isosceles.
Solution:
A (6, 10, 10), B (1, 0, -5), C (6, -10, 0) are the vertices of ∆ABC
D.Rs of AB are 5, 10, 15 i.e., 1, 2, 3
D.Rs of BC are -5, 10, -5 i.e., 1, -2, 1
D.Rs of AC are 0,20,10, i.e., 0, 2, 1
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 13
∴ The given triangle is right angle a.

Question 7.
The vertices of a triangle are A (1, 4, 2), B (-2, 1, 2) C (2, 3, -4). Find ∠A, ∠B, ∠C.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 14
Solution:
A (1, 4, 2), B (-2, 1, 2), C (2, 3, -4) and the vertices of OABC.
D.Rs of AB are 3, 3, 0 i.e., 1, 1, 0
D.Rs of BC are -4, -2, 6 i.e., 2, 1, -3
D.Rs of AC are -1, 1, 6.
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 15
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 16

Question 8.
Find the angle between the lines whose direction cosines are given by the equation 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0
Solution:
Given 3l + m + 5n = 0 ……….. (1)
6mn – 2nl + 5lm = 0 …………. (2)
From (1), m = – (3l + 5n)
Substituting in (2)
-6n(3l + 5n) -2nl – 5l(3l + 5n) = 0
-18ln – 30n² – 2nl – 15l² – 25ln = 0
-15l² – 45ln – 30n² = 0
l² + 3ln + 2n² = 0
(l + 2n) (l + n) — 0
l + 2n = 0 or l + n = 0

Case (i) : l1 + n1 = 0 ⇒ n1 = – l1; ⇒ \(\frac{l_{1}}{1}=\frac{n_{1}}{-1}\)
But m1 = -(3l1 + 5n1) = -(-3n1 + 5n1) = -2n1
∴ \(\frac{m_{1}}{+2}=\frac{n_{1}}{-1}\)
∴ \(\frac{l_{1}}{1}=\frac{m_{1}}{2}=\frac{n_{1}}{-1}\)
D.Rs of the first line lx are (1,2, -1)
Case (ii): l2 + 2n2 = 0
l2 = —2n2 ⇒ \(\frac{l_{2}}{-2}=\frac{n_{2}}{1}\)
m2 = -(3l2 + 5n2) = – (-6n2 + 5n2) = n2
\(\frac{m_{2}}{1}=\frac{n_{2}}{1}\)
∴ \(\frac{l_{2}}{-2}=\frac{m_{2}}{1}=\frac{n_{2}}{1}\)
D.Rs of the second line Z2 are (-2, 1, 1)
Suppose ‘θ’ is the angle between the lines l1 and l2
Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b) 17

Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Ex 6(b)

Question 9.
If a variable line in two adjecent positions has direction cosines (l, m, n) and (l + δl, m + δm, n + δn), show that the small angle δθ between two position. Is given by (δθ)² = (δl)² + (δm)² + (δn)².
Solution:
Since (l, m, n) and (l + δl, m + δm, n + δn) are direction cosines, we have
l² + m² + n² = 1 …………. (1)
(l + δl)² + (m + δm)² + (n + δn)² — 1 ………. (2)
(2) – (1) gives (l + δl)² + (m + δm)² + (n + 8n)² – (l² + m² + n²) = 0
2(l.δl + m.δm + n.δn) = -((δl)² + (δm)² + (δn)²) …………. (3)
cos θ . δθ = l(l + δl) + m(m + δm) + n(n + δn)
= (l² + m² + n²) + (l.δl + m.δm + n.δn)
= 1 – \(\frac{1}{2}\) [(δl)² + (δm)² + (δn)²]
(δl)² + (δm)² + (δn)² = 1 = ² (1 – cos θ.δ θ)
δθ being small, sin \(\frac{\delta \theta}{2}\) = \(\frac{\delta \theta}{2}\)
∴ sin²θ \(\frac{\delta \theta}{2}\) = 4(\(\frac{\delta \theta}{2}\))² = (δθ)²
∴ (δθ)² = (δl)² + (δm)² + (δn)².