The two lines x = ay + b , z = cy + d ; and x = a ' y + b' , z = c ' y + d ' are perpendicular to each other if

Question

The two lines x = ay + b , z = cy + d ; and x = a ' y + b' , z = c ' y + d ' are perpendicular to each other if (A) aa' + cc' = -1 (B) aa' + cc' = 1 (C) (a/a') + (c/c') = - 1  (D) (a/a') + (c/c') = 1

Tardigrade Admin · Accepted Answer

Equation of lines (x-b/a) = (y/1) = (z-d/c) (x-b'/a') = (y/1) = (z-d'/c') Line are perpendicular ⇒ aa' + 1+ cc' = 0

Q. The two lines $x = a y + b, z = cy + d;$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

aa' + cc' = -1

aa' + cc' = 1

$\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = - 1$

$\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = 1$

Equation of lines $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$
$\frac{x - b ^{'}}{a ^{'}} = \frac{y}{1} = \frac{z - d ^{'}}{c ^{'}}$
Line are perpendicular
$\Rightarrow a a^{'} + 1 + c c^{'} = 0$

Q. The two lines x=ay+b,z=cy+d; and x=a′y+b′,z=c′y+d′ are perpendicular to each other if

aa' + cc' = -1

aa' + cc' = 1

a′a​+c′c​=−1

a′a​+c′c​=1

Equation of lines ax−b​=1y​=cz−d​ a′x−b′​=1y​=c′z−d′​ Line are perpendicular ⇒aa′+1+cc′=0

Q. The two lines $x = a y + b, z = cy + d;$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

$\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = - 1$

$\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = 1$

Equation of lines $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$
$\frac{x - b ^{'}}{a ^{'}} = \frac{y}{1} = \frac{z - d ^{'}}{c ^{'}}$
Line are perpendicular
$\Rightarrow a a^{'} + 1 + c c^{'} = 0$