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Question 47 of 59

Q.The angle between the straight lines (x-4)/2 = (y-5)/0 = (z-6)/0 and (3-x)/3 = (y-7)/0 = (z-3)/0 is

(a) -π
(b) -π/2
(c) π
(d) π/3
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2023MCQ· 1mImportance★★★★★
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The angle between two lines is found from the angle between their direction vectors using cos⁡θ=∣d⃗1⋅d⃗2∣∣d⃗1∣∣d⃗2∣\cos\theta=\dfrac{|\vec d_1\cdot\vec d_2|}{|\vec d_1||\vec d_2|} — but here the sign of the dot product itself tells us the lines point in exactly opposite directions.

Step 1. First line: x−42=y−50=z−60\dfrac{x-4}{2}=\dfrac{y-5}{0}=\dfrac{z-6}{0} has direction ratios (2,0,0)(2,0,0).

Step 2. Second line: 3−x3=y−70=z−30\dfrac{3-x}{3}=\dfrac{y-7}{0}=\dfrac{z-3}{0}, i.e. x−3−3=y−70=z−30\dfrac{x-3}{-3}=\dfrac{y-7}{0}=\dfrac{z-3}{0}, has direction ratios (−3,0,0)(-3,0,0).

Step 3. cos⁡θ=(2)(−3)+0+04⋅9=−66=−1⇒θ=π\cos\theta = \dfrac{(2)(-3)+0+0}{\sqrt{4}\cdot\sqrt{9}} = \dfrac{-6}{6}=-1 \Rightarrow \theta=\pi. …

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