The angle between the straight lines x + $\frac{1}{2}$ = y - $\frac{2}{5}$ = z + $\frac{3}{4}$ and x - $\frac{1}{1}$ = y + $\frac{2}{2}$ = z - 3/- 3 is

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(d) $90^\circ$ We have , x + $\frac{1}{2}$ = y - $\frac{2}{5}$ = z + $\frac{3}{4}$ x - $\frac{1}{1}$ = y + $\frac{2}{2}$ = z - 3/- 3 The direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, -3. The given lines are parallel to the vectors \overrightarrow{b_1} = 2 \hat{i} + 5 \hat{j} + 4 \hat{k} \text{ and } \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k} Let $\theta$ be the angle between the given lines. Now, \cos $\theta$ = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|} = \frac{\left( 2 \hat{i} + 5 \hat{j} + 4 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} - 3 \hat{k} \right)}{$\sqrt$(2^2 + 5^2 + 4^2) \sqrt{1^2 + 2^2 + \left( - 3 \right)^2}} = 2 + 10 - 12/$\sqrt$(45) $\sqrt$(14) = 0 \Rightarrow $\theta$ = $90^\circ$

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