If the lines x = ay + b, z = cy + d and x = a | vedclass

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(B) The first line is given by $x = ay + b$ and $z = cy + d$. This can be written in symmetric form as $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c

Vedclass · Accepted Answer

$aa' + c + c' = 0$

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(B) The first line is given by $x = ay + b$ and $z = cy + d$. This can be written in symmetric form as $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$. The direction ratios of this line are $(a, 1, c)$. The second line is given by $x = a'z + b'$ and $y = c'z + d'$. This can be written in symmetric form as $\frac{x - b'}{a'} = \frac{y - d'}{c'} = \frac{z}{1}$. The direction ratios of this line are $(a', c', 1)$. Since the two lines are perpendicular,the dot product of their direction ratios must be zero: $(a)(a') + (1)(c') + (c)(1) = 0$. Therefore,$aa' + c' + c = 0$.

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

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