The lines frac{1-x}{3} = frac{7y-14}{2p} = fr | vedclass

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

Vedclass · Accepted Answer

$\frac{70}{11}$

Vedclass · Answer

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

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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