Find the vector and Cartesian equations of the line passing through the point (1,2,-4) and perpendicular to the two lines x-1=y+19=z and x=y-98=z-99.

Let the Cartesian equation of line passing through (1,2,-4) be x-1/a = y-2/b = z+4/c — (1)
Given lines are x-1=y+19=z —(2)
x=y-98=z-99 —(3)
Obviously parallel vectors →b1, →b2, →b3 of (1), (2), (3) respectively are given as
→b1 = a→i+b→j+c→k
→b2 = →i-→j+→k
→b3 = →i+→j+→k
As per the question,(1)⊥(2) ⇒ →b1⊥→b2 ⇒ →b1.→b2=0
(1)⊥(3) ⇒ →b1⊥→b3 ⇒ →b1.→b3=0
Hence, a-b+c=0 — (4)
and a+b+c=0 — (5)
From equtaion (4) and (5), a=0, b=c
⇒a/0 = b/1 = c/1 ⇒ a=0, b=λ, c=λ
Putting the value of a, b and c in (1) we get required cartesian equation of line as
x-1/0 = y-2/λ = z+4/λ ⇒ x-1=0 , y-2 = z+4
Hence x=1 , y-z=6
Hence vector equation is →r=(→i+2→j-4→k)+λ(→j+→k).