If the lines frac{1-x}{3}=frac{7y-14}{2p}=fra | vedclass

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(A) First,rewrite the equations of the lines in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.For the first line: $\frac{x-1}{-3

Vedclass · Accepted Answer

$\frac{70}{11}$

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(A) First,rewrite the equations of the lines in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.For the first line: $\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$. The direction ratios are $\vec{v_1} = (-3, \frac{2p}{7}, 2)$.For the second line: $\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}$. The direction ratios are $\vec{v_2} = (-\frac{3p}{7}, 1, -5)$.Since the lines are at right angles,their dot product must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$.$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.$\frac{11p}{7} = 10$.$p = \frac{70}{11}$.

If 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,then $p=$

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