Question:
Let \( \alpha, \beta \) and \( \gamma \) be the angles made by a straight line with the x-axis, y-axis and z-axis respectively. If \( \cos\alpha + \cos\beta + \cos\gamma = \frac{5}{3} \), then the value of \( \cos\alpha \cos\beta + \cos\beta \cos\gamma + \cos\gamma \cos\alpha \) is equal to
Let \( \alpha, \beta \) and \( \gamma \) be the angles made by a straight line with the x-axis, y-axis and z-axis respectively. If \( \cos\alpha + \cos\beta + \cos\gamma = \frac{5}{3} \), then the value of \( \cos\alpha \cos\beta + \cos\beta \cos\gamma + \cos\gamma \cos\alpha \) is equal to
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Always use \( l^2+m^2+n^2=1 \) with \( (l+m+n)^2 \) identity for such problems.
Updated On: Apr 21, 2026
- \( \frac{11}{3} \)
- \( \frac{8}{9} \)
- \( \frac{11}{9} \)
- \( \frac{7}{3} \)
- \( \frac{7}{9} \)
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The Correct Option is B
Solution and Explanation
Concept:
Direction cosines satisfy:
\[
l^2 + m^2 + n^2 = 1
\]
Step 1: Let \( l = \cos\alpha, m = \cos\beta, n = \cos\gamma \).
\[ l + m + n = \frac{5}{3} \]
Step 2: Use identity.
\[ (l+m+n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl) \] \[ \left(\frac{5}{3}\right)^2 = 1 + 2(lm + mn + nl) \]
Step 3: Solve.
\[ \frac{25}{9} = 1 + 2S \Rightarrow \frac{25}{9} - 1 = 2S \] \[ \frac{16}{9} = 2S \Rightarrow S = \frac{8}{9} \]
Step 1: Let \( l = \cos\alpha, m = \cos\beta, n = \cos\gamma \).
\[ l + m + n = \frac{5}{3} \]
Step 2: Use identity.
\[ (l+m+n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl) \] \[ \left(\frac{5}{3}\right)^2 = 1 + 2(lm + mn + nl) \]
Step 3: Solve.
\[ \frac{25}{9} = 1 + 2S \Rightarrow \frac{25}{9} - 1 = 2S \] \[ \frac{16}{9} = 2S \Rightarrow S = \frac{8}{9} \]
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