Question

If the direction ratios of two lines are \( (3,0,2) \) and \( (0,2,k) \), and \( \theta \) is the angle between them, and if \( |\cos \theta| = \frac{6}{13} \), then \( k = \)

• A. \( \pm 2 \)
• B. \( \pm 3 \)
• C. \( \pm 5 \)
• D. \( \pm 7 \)

Hint:

For finding the angle between two lines using direction ratios, use the formula \( \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}} \).
- The magnitude of the cosine value can help to simplify the equation for \( k \).

Answer 1

The Correct Option is B

The formula for the cosine of the angle between two vectors is given by: \[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}} \] Here, the direction ratios of the two lines are \( (3,0,2) \) and \( (0,2,k) \). Using the formula, we have: \[ |\cos \theta| = \frac{3(0) + 0(2) + 2(k)}{\sqrt{3^2 + 0^2 + 2^2} \cdot \sqrt{0^2 + 2^2 + k^2}} = \frac{2k}{\sqrt{9 + 4} \cdot \sqrt{4 + k^2}}. \] Simplifying, we get: \[ \frac{2k}{\sqrt{13} \cdot \sqrt{4 + k^2}} = \frac{6}{13}. \] Squaring both sides: \[ \frac{4k^2}{13(4 + k^2)} = \frac{36}{169}. \] Cross-multiply to solve for \( k \), and we obtain: \[ 4k^2 = \frac{36}{169} \times 13(4 + k^2), \] which simplifies to \( k = \pm 3 \). Thus, the correct answer is \( \boxed{k = \pm 3} \).