Question

If the angle between the lines whose direction ratios are \( 4, -3, 5 \) and \( 3, 4, k \) is \( \frac{\pi}{3} \), then \( k = \)

• A. \( \pm 7 \)
• B. \( \pm 10 \)
• C. \( \pm 5 \)
• D. \( \pm 6 \)

Hint:

To find the angle between two lines, use the formula for the dot product of their direction ratios. The cosine of the angle is the ratio of this dot product to the product of their magnitudes.

Answer 1

The Correct Option is C

Step 1: Formula for the angle between two lines.
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: \[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \] We are given that \( \theta = \frac{\pi}{3} \), so \( \cos \frac{\pi}{3} = \frac{1}{2} \). Step 2: Substituting the values.
The direction ratios of the lines are \( (4, -3, 5) \) and \( (3, 4, k) \). Substituting these into the formula: \[ \frac{4 \cdot 3 + (-3) \cdot 4 + 5 \cdot k}{\sqrt{4^2 + (-3)^2 + 5^2} \sqrt{3^2 + 4^2 + k^2}} = \frac{1}{2} \] Simplifying the numerator and denominator: \[ \frac{12 - 12 + 5k}{\sqrt{16 + 9 + 25} \sqrt{9 + 16 + k^2}} = \frac{1}{2} \] \[ \frac{5k}{\sqrt{50} \sqrt{25 + k^2}} = \frac{1}{2} \] Step 3: Solving for \( k \).
Cross-multiply and solve for \( k \): \[ 10k = \sqrt{50} \sqrt{25 + k^2} \] Squaring both sides: \[ 100k^2 = 50(25 + k^2) \] \[ 100k^2 = 1250 + 50k^2 \] \[ 50k^2 = 1250 \] \[ k^2 = 25 \] Thus, \( k = \pm 5 \). Step 4: Conclusion.
Thus, the value of \( k \) is \( \boxed{\pm 5} \).