From a point P λ , λ , λ , perpendiculars PQ and PR are drawn respectively on the lines y = x , z = 1 and y = - x , z = - 1 . If P is such that ∠ Q P R is a right angle, then the possible value (s) of λ is (are)

Step-by-step solution

Line 1 : x 1 = y 1 = z - 1 0 - r , Q r , r , 1 Line 2 : x 1 = y - 1 = z + 1 0 = k , R k , - k , - 1 P Q → = λ - r i ^ + λ - r j ^ + λ - 1 k ^ And λ - r + λ - r = 0 as P Q → is ⊥ to L 1 ⇒ 2 λ = 2 r ⇒ λ = r P R → = λ - k i ^ + λ + k j ^ + λ + 1 k ^ And λ - k - λ - k = 0 as P R → is ⊥ to L 2 ⇒ k = 0 So P Q ⊥ P R λ - r λ - k + λ - r λ + k + λ - 1 λ + 1 = 0 ⇒ λ = 1 , - 1 For λ = 1 as point P and Q coincide ⇒ λ = - 1 .