The lines x = ay + b, z = cy + d and x = a'y | vedclass

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(B) The given lines can be written in symmetric form as follows:For the first line $x = ay + b$ and $z = cy + d$,we have $\frac{x - b}{a} = y = \frac{

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$aa' + cc' = -1$

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(B) The given lines can be written in symmetric form as follows:For the first line $x = ay + b$ and $z = cy + d$,we have $\frac{x - b}{a} = y = \frac{z - d}{c}$.Thus,the direction ratios of the first line are $(a, 1, c)$.For the second line $x = a'y + b'$ and $z = c'y + d'$,we have $\frac{x - b'}{a'} = y = \frac{z - d'}{c'}$.Thus,the direction ratios of the second line are $(a', 1, c')$.Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Substituting the direction ratios,we get $(a)(a') + (1)(1) + (c)(c') = 0$.Therefore,$aa' + 1 + cc' = 0$,which implies $aa' + cc' = -1$.

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

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