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Mathematics · JEE | THREE DIMENSIONAL GEOMETRY
हिन्दीIf the lines \( \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2} \) and \( \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \) are perpendicular to each other, then the value of \( p \) is:
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Correct answer · Option B
\( \frac{70}{11} \)
Explanation
Standard form: \( \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2} \) and \( \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5} \). Perpendicularity condition: \( (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0 \implies \frac{9p}{7} + \frac{2p}{7} - 10 = 0 \implies \frac{11p}{7} = 10 \implies p = \frac{70}{11} \).
