→ (t), → (p, r), → (q) (JEE given q, s) → (r)

Sol. Let the line through origin is = =

⇒ x = λ z , y = µz ...........

To find point of intersection of line and line = = ..........

we have = = z + 1

⇒ z = =

⇒ λ + 3µ + 5 = 0 ..........

To find point of intersection of line and line = = ........

we have = =

⇒ z = =

⇒ 3 λ + µ = 5 ............(5)

Solving and (5), λ = and µ =

∴ z = 2, x = 5, y = – 5 for point P

and z = , x = , y = for point Q

∴ PQ 2 = + + = 6

tan –1 (x + 3) – tan –1 (x – 3) = sin –1 (3/5)

⇒ tan –1 = tan –1 (3/4)

⇒ = ⇒ x 2 = 16

∴ x = ± 4

Since . = 0

∴ Let = λ 1 , = λ 2

Now 2| + | = | – | & = µ + 4 ⇒ =

= 2

squaring

= ⇒ = (12 + µ 2 – 8µ) .........

Also ( – ).( + ) = 0

⇒ . = 0 ⇒ = 0

⇒ ..............

from &

12 + µ 2 – 8µ = 12 – 3µ

⇒ µ 2 – 5µ = 0 ⇒ µ = 0, 5 (µ = 5 reject)

Ι = = = dx

Ι = dx ......(i)

(using = )

= dx ......(ii)

Add (i) and (ii)

Ι = dx

Consider

Ι k – Ι k–2 = dx = dx

Ι k = Ι k–2

so Ι 5 = Ι 3 ⇒ Ι 5 = Ι 1 = = 4

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