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COMEDK · Maths · 34. Three Dimensional Geometry

Consider two skew lines in 3D space.
\(M_1: \dfrac{x-1}{1} = \dfrac{2-y}{1} = \dfrac{z-5}{1}\) and \(M_2: \dfrac{x+3}{1} = \dfrac{y-7}{2} = \dfrac{z+4}{1}\)
Let \(L_1\) be the line of shortest distance (common perpendicular) between \(M_1\) and \(M_2\)
If \(L_2\) is a line parallel to the vector \(\vec{b} = \hat{j} + \hat{k}\),
Then the acute angle \(\theta\) between the lines \(L_1\) and \(L_2\) is:

  1. A \(30^\circ\)
  2. B \(\cos^{-1}\left(\dfrac{1}{3}\right)\)
  3. C \(45^\circ\)
  4. D \(60^\circ\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(60^\circ\)

Step-by-step Solution

Detailed explanation

The equation of line \(M_1\) can be rewritten in standard form as: \(\dfrac{x-1}{1} = \dfrac{y-2}{-1} = \dfrac{z-5}{1}\) The direction vector of \(M_1\) is \(\vec{v}_1 = \hat{i} - \hat{j} + \hat{k}\). The direction vector of \(M_2\) is…