TS Inter 1st Year Maths 1B Solutions Chapter 6 Direction Cosines and Direction Ratios Ex 6(a)

Students must practice these TS Intermediate Maths 1B Solutions Chapter 6 Direction Cosines and Direction Ratios Ex 6(a) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Direction Cosines and Direction Ratios 6(a)

I.
Question 1.
A line makes angles 90°, 60° and 30° with positive directions of X, Y, Z axes respectively. Find the direction cosines. (V.S.A.Q.)
Answer:
If l, m, n are the d.c.’s of the line
l = cos l = cos 90° = 0, m = cos β = cos 60° = \(\frac{1}{2}\)
n = cos γ = cos 30° = \(\frac{\sqrt{3}}{2}\)
d.c.’s of the line are \(\left(0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)

Question 2.
If a line makes angles α, β, γ with the positive directions of X, Y, Z axes, what is the value of sin²α + sin²β + sin²γ ? (V.S.A.Q.)
Answer:
sin²α + sin²β + sin²γ
= 1 – cos²α + 1 – cos²β + 1 – cos²γ
= 3 – l² – m² – n²
= 3 – (l² + m² + n²) = 3 – 1 = 2
(cos α = l, cos β = m, cos γ = n are d.c.’s of a line)

Question 3.
If P (√3 , 1, 2√3) is a point in space, find the direction cosines of OP (V.S.A.Q.)
Answer:
Direction ratios of
\(\overrightarrow{\mathrm{OP}}\) = (√3 – 0, 1 – 0, 2√3 – 0) = (√3, 1, 2√3)
∴ a2+ b2 + c2 = 3 + 1+ 12 = 16
⇒ \(\sqrt{a^2+b^2+c^2}\) = 4
∴ Direction cosines of \(\overrightarrow{\mathrm{OP}}\)

Question 4.
Find the direction cosines of the line joining the points (-4, 1, 7) and (2, -3, 2). (V.S.A.Q.)
Answer:
Let A = (-4, 1, 7) and B = (2, -3, 2)
d.r.’s of AB = (2 + 4, -3 -1, 2 – 7) = (6, -4, -5)

II.
Question 1.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2). (S.A.Q.)
Answer:
Let A (3, 5, -4), B (-1, 1, 2) and C (-5, -5, -2) be the vertices of ∆ABC.
d.r.’s of AB = (-1 – 3, 1 – 5, 2 + 4) = (-4, -4, 6)

Question 2.
Show that the lines \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{RS}}\) are parallel where P, Q, R, S erne the points (2, 3, 4), (4, 7, 8), (-1, -2, 1) and (1, 2, 5) respectively. (V.S.A.Q.)
Answer:
P (2, 3, 4), Q ( 4, 7, 8), R (-1, -2, 1) and S (1, 2, 5) are the given points.
d.r.’s of \(\overleftrightarrow{\mathrm{PQ}}\) = (4 — 2,7 — 3, 8 – 4) = (2, 4, 4)
d.r.’s of \(\overleftrightarrow{\mathrm{RS}}\) = (1 + 1 , 2 + 2, 5 – 1) = (2, 4, 4)
d.r.’s of \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{RS}}\) are proportional. Hence \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{RS}}\) are parallel.

III.
Question 1.
Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l² + 5m² – 3n² = 0 (E.Q.) (June 2009)
Answer:
The given relations are
l – 5m + 3n = 0 …………………….. (1)
7l² + 5m² – 3n² = 0 ………………….. (2)
From (1), l = 5m – 3n ………………….. (3)
∴ From (2),
7(5m – 3n)² + 5m² – 3n² = 0
⇒ 7(25m² – 30mn + 9n²) + 5m² – 3n² = 0
⇒ 175m² + 63n² – 210mn + 5m² – 3n² = 0
⇒ 180m² – 210mn + 60n² = 0
⇒ 6m² – 7mn + 2n² = 0
⇒ (3m – 2n) (2m – n) = 0
⇒ 3m = 2n ⇒ m = \(\frac{2 n}{3}\) (or) 2m = n ⇒ m = \(\frac{n}{2}\)

Case (i):

Case (ii):