Find the values of p so that the lines frac{1 | vedclass

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(A) The given equations can be rewritten in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ as follows:For the first line: $\f

Vedclass · Accepted Answer

$p = \frac{70}{11}$

Vedclass · Answer

(A) The given equations can be rewritten in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ as follows:For the first line: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$.For the second line: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$.The direction ratios of the lines are $a_1 = -3, b_1 = \frac{2p}{7}, c_1 = 2$ and $a_2 = \frac{-3p}{7}, b_2 = 1, c_2 = -5$.Two lines are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.Substituting the values: $(-3)\left(\frac{-3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0$.$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.$\frac{11p}{7} = 10$.$11p = 70$.$p = \frac{70}{11}$.

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.

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