Question

Let \( \alpha, \beta, \gamma \) be the direction cosines of a vector \( \vec{a} = x \hat{i} + y \hat{j} + z \hat{k} \), where \( z<0 \). If \( \alpha = \frac{-4}{\sqrt{105}} \) and \( \beta = \frac{\sqrt{5}}{\sqrt{21}} \), then \( \gamma \) is equal to

• A. \( \frac{-8}{\sqrt{105}} \)
• B. \( \frac{-\sqrt{8}}{\sqrt{105}} \)
• C. \( \frac{-5}{\sqrt{105}} \)
• D. \( \frac{-5}{\sqrt{21}} \)
• E. \( \frac{-8}{\sqrt{21}} \)

Hint:

The sum of squares of direction cosines is always 1: \[ \alpha^2 + \beta^2 + \gamma^2 = 1 \]

Answer 1

The Correct Option is A

The direction cosines \( \alpha, \beta, \gamma \) satisfy the equation: \[ \alpha^2 + \beta^2 + \gamma^2 = 1 \] Substituting given values: \[ \left( \frac{-4}{\sqrt{105}} \right)^2 + \left( \frac{\sqrt{5}}{\sqrt{21}} \right)^2 + \gamma^2 = 1 \] Calculating each term: \[ \frac{16}{105} + \frac{5}{21} + \gamma^2 = 1 \] Converting \( \frac{5}{21} \) to denominator 105: \[ \frac{16}{105} + \frac{25}{105} + \gamma^2 = 1 \] \[ \frac{41}{105} + \gamma^2 = 1 \] \[ \gamma^2 = 1 - \frac{41}{105} \] \[ \gamma^2 = \frac{105}{105} - \frac{41}{105} = \frac{64}{105} \] \[ \gamma = \pm \frac{8}{\sqrt{105}} \] Since \( z<0 \), we take the negative value: \[ \gamma = \frac{-8}{\sqrt{105}} \] 
Final Answer: \[ \boxed{\frac{-8}{\sqrt{105}}} \]