If the lines x = ay + b, z = cy + d and x= a'z + b', y = c'z + d' are perpendicular, then:

Question

If the lines x = ay + b, z = cy + d and x= a'z + b', y = c'z + d' are perpendicular, then: (A) cc' + a + a' = 0 (B) aa' + c + c' = 0 (C) ab' + bc' + 1 = 0 (D) bb' + cc' + 1 = 0

Tardigrade Admin · Accepted Answer

Line x = ay + b, z = cy + d ⇒ (x-b/a) = (y/1) = (z-d/c) Line x =a'z + b', y =c'z +d' ⇒ (x-b'/a') = (y-d'/c') =(z/1) Given both the lines are perpendicular ⇒ aa' +c' +c =0

Q. If the lines $x = a y + b, z = cy + d$ and $x = a^{'} z + b^{'}, y = c^{'} z + d^{'}$ are perpendicular, then:

$c c^{'} + a + a^{'} = 0$

$a a^{'} + c + c^{'} = 0$

$a b^{'} + b c^{'} + 1 = 0$

$b b^{'} + c c^{'} + 1 = 0$

Line $x = a y + b, z = cy + d \Rightarrow \frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$
Line $x = a^{'} z + b^{'}, y = c^{'} z + d^{'} \Rightarrow \frac{x - b ^{'}}{a ^{'}} = \frac{y - d ^{'}}{c ^{'}} = \frac{z}{1}$
Given both the lines are perpendicular
$\Rightarrow a a^{'} + c^{'} + c = 0$

Q. If the lines x=ay+b,z=cy+d and x=a′z+b′,y=c′z+d′ are perpendicular, then:

cc′+a+a′=0

aa′+c+c′=0

ab′+bc′+1=0

bb′+cc′+1=0