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(A) Let $\bar{b}_{1}$ and $\bar{b}_{2}$ be the vectors parallel to the given lines.The direction ratios of the first line are $(2, 5, -3)$,so $\bar{b}

Vedclass · Accepted Answer

$Q=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)$

Vedclass · Answer

(A) Let $\bar{b}_{1}$ and $\bar{b}_{2}$ be the vectors parallel to the given lines.The direction ratios of the first line are $(2, 5, -3)$,so $\bar{b}_{1} = 2 \hat{i} + 5 \hat{j} - 3 \hat{k}$.The direction ratios of the second line are $(-1, 8, 4)$,so $\bar{b}_{2} = -\hat{i} + 8 \hat{j} + 4 \hat{k}$.Calculate the magnitudes:$|\bar{b}_{1}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.$|\bar{b}_{2}| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$.Calculate the dot product:$\bar{b}_{1} \cdot \bar{b}_{2} = (2)(-1) + (5)(8) + (-3)(4) = -2 + 40 - 12 = 26$.The angle $Q$ between the lines is given by:$\cos Q = \frac{|\bar{b}_{1} \cdot \bar{b}_{2}|}{|\bar{b}_{1}| |\bar{b}_{2}|} = \frac{|26|}{9 \sqrt{38}} = \frac{26}{9 \sqrt{38}}$.Therefore,$Q = \cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)$.

Find the angle between the following pair of lines:
$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

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Find the angle between the following pair of lines:$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$