Question

The direction ratios of a straight line \(L_1\) are 2,-1,2 and that of another straight line \(L_2\) are 3,6,-2. Then the angle between \(L_1\) and \(L_2\) is

• A. \(\cos^{-1}\left(\frac{-4}{21}\right)\)
• B. \(\cos^{-1}\left(\frac{8}{21}\right)\)
• C. \(\cos^{-1}\left(\frac{5}{21}\right)\)
• D. \(\cos^{-1}\left(\frac{-4}{7}\right)\)
• E. \(\cos^{-1}\left(\frac{10}{21}\right)\)

Hint:

If you need to find whether lines are perpendicular, just check if the numerator \(a_1 a_2 + b_1 b_2 + c_1 c_2\) is zero. This can save you from calculating square roots in many cases.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The cosine of the angle \(\theta\) between two lines with direction ratios \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) is given by the dot product formula.

Step 2: Key Formula or Approach:
\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Step 3: Detailed Explanation:
Given DRs: \(L_1 = (2, -1, 2)\) and \(L_2 = (3, 6, -2)\). \[ \text{Numerator} = (2)(3) + (-1)(6) + (2)(-2) = 6 - 6 - 4 = -4 \]
\[ \text{Magnitude of } L_1 = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \]
\[ \text{Magnitude of } L_2 = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9+36+4} = \sqrt{49} = 7 \]
Substituting into the formula: \[ \cos \theta = \frac{-4}{3 \times 7} = \frac{-4}{21} \]
\[ \theta = \cos^{-1}\left(\frac{-4}{21}\right) \]

Step 4: Final Answer:
The angle is \(\cos^{-1}\left(\frac{-4}{21}\right)\).