If the angle between the lines,frac{x}{2} = f | vedclass

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(D) The direction ratios of the first line $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ are $\vec{v_1} = (2, 2, 1)$.The second line is $\frac{-(x - 5)}{-

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$\frac{7}{2}$

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(D) The direction ratios of the first line $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ are $\vec{v_1} = (2, 2, 1)$.The second line is $\frac{-(x - 5)}{-2} = \frac{7(y - 2)}{p} = \frac{z - 3}{4}$,which simplifies to $\frac{x - 5}{2} = \frac{y - 2}{p/7} = \frac{z - 3}{4}$.The direction ratios of the second line are $\vec{v_2} = (2, p/7, 4)$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.Given $\cos 	heta = \frac{2}{3}$,we have $\frac{2}{3} = \frac{|2(2) + 2(p/7) + 1(4)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{2^2 + (p/7)^2 + 4^2}}$.$\frac{2}{3} = \frac{|8 + 2p/7|}{3 \sqrt{4 + p^2/49 + 16}} = \frac{|8 + 2p/7|}{3 \sqrt{20 + p^2/49}}$.Canceling $3$ from the denominator,we get $\frac{2}{1} = \frac{|8 + 2p/7|}{\sqrt{20 + p^2/49}}$.Squaring both sides: $4 = \frac{(8 + 2p/7)^2}{20 + p^2/49} = \frac{64 + 32p/7 + 4p^2/49}{20 + p^2/49}$.$80 + 4p^2/49 = 64 + 32p/7 + 4p^2/49$.$80 = 64 + 32p/7 \Rightarrow 16 = 32p/7 \Rightarrow p = \frac{16 	imes 7}{32} = \frac{7}{2}$.

If the angle between the lines,$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7y - 14}{p} = \frac{z - 3}{4}$ is $\cos^{-1} \left( \frac{2}{3} \right)$,then $p$ is equal to

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