The two lines $x=a y+b, z=c y+d$ and $x=a^{\prime} y+b^{\prime} z=c^{\prime} y+d^{\prime}$ will be…
The two lines $x=a y+b, z=c y+d$ and $x=a^{\prime} y+b^{\prime} z=c^{\prime} y+d^{\prime}$ will be perpendicular, if and only if
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$a a^{\prime}+c c^{\prime}+1=0$
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$\mathrm{a} \mathrm{a}^{\prime}+\mathrm{b} \mathrm{b}^{\prime}+\mathrm{c} \mathrm{c}^{\prime}+1=0$
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$\mathrm{a}^{\prime}+\mathrm{b} \mathrm{b}^{\prime}+\mathrm{c} \mathrm{c}^{\prime}=0$
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$\left(a+a^{\prime}\right)\left(b+b^{\prime}\right)+\left(c+c^{\prime}\right)=0$
Solution
$\frac{\mathrm{x}-\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{y}}{1}=\frac{3-\mathrm{d}}{\mathrm{c}} ; \frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{y}}{1}=\frac{3-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}$
For perpendicular $\mathrm{aa}^{\prime}+1+\mathrm{cc}^{\prime}=0$
Asked in: JEE Main 2003
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