and it intersects the lines. ⇒ S.D. = 0 ⇒ a + 3b + 5c = 0 and 3a + b - 5c = 0 ⇒ a : b : c :: 5r : - 5r : 2r On solving with given lines, we get points of intersection: P = (5 , - 5 , 2) and Q = (
10
3
,−
10
3
,
8
3
) ⇒ PQ2 = d2 = 6. (B)→(P), (R) We have tan−1(x+3)−tan−1(x−3) = sin−1(
3
5
) ⇒ tan−1
(x+3)−(x−3)
1+(x2−9)
= tan−1
3
4
⇒
6
x2−8
=
3
4
Hence, x2−8 = 8 or x = ± 4. (C)→(Q), (S) As
→
a
= µ
→
b
+4
→
c
we have µ(|
→
b
|) = −4
→
b
.
→
c
Also, |
→
b
|2 = 4
→
a
.
→
c
and |
→
b
|+
→
b
.
→
c
−
→
d
.
→
c
= 0 Also, 2|
→
b
+
→
c
| = |
→
b
−
→
a
|. Solving and eliminating
→
b
.
→
c
and eliminating |
→
a
|, we get (2µ2−10µ)|
→
b
|2 = 0 Therefore, µ = 0 and 5. (D)→(R) We have I =
2
π
π
∫
−π
sin9(
x
2
)
sin(x−2)
dx =
2
π
×2
π
∫
0
sin9(
x
2
)
sin(
x
2
)
dx
x
2
= θ ⇒ dx = 2dθ When x = 0 , θ = 0 When x = π , θ = π/2 Therefore, I =