Find the values of p so that the lines ...

NCERT Class XII Mathematics Chapter - - Solutions

Question : 19 of 78

Marks: +1, -0

Find the values of p so that 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 at right angles.

Solution:

Equations of the given lines can be written as

\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}

and

\frac{x-1}{-\frac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}

Direction ratios of these lines are respectively

\langle -3, \frac{2p}{7}, 2 \rangle

and

\langle \frac{-3p}{7}, 1, -5 \rangle

The lines are at right angles if

(-3)\times\left(-\frac{3p}{7}\right)

+\left(\frac{2p}{7}\right)\times 1 + 2\times (-5) = 0

\Rightarrow \frac{11p}{7} = 10

i.e., if

p = \frac{70}{11}

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