$f$ the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$ are perpendicular, then
Select the correct option:
A
$a b^{\prime}+b c^{\prime}+1=0$
B
$c c^{\prime}+a+a^{\prime}=0$
C
$a a^{\prime}+c+c \prime=0$
D
$b b^{\prime}+c c^{\prime}+1=0$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
irst line is : $x=a y+b, z=c y+d$
$$
\Rightarrow \frac{x-b}{a}=y=\frac{z-d}{c}
$$
and another line is: $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$
$$
=\frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{z}-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}=\mathrm{z}
$$
∵ both lines are perpendicular to each other
$$
\therefore a a^{\prime}+c^{\prime}+c=0
$$
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