MHT CET · Maths · Three Dimensional Geometry
The direction co-sines of a line which makes equal acute angles with the co-ordinate
axes are
- A \(\frac{-1}{3}, \frac{1}{3}, \frac{1}{3}\)
- B \(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\)
- C \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
- D \(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
We know that
\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) and given \(\alpha=\beta=\gamma\)
\(\therefore 3 \cos ^{2} \alpha=1 \Rightarrow \cos ^{2} \alpha=\frac{1}{3} \Rightarrow \cos \alpha=\cos \beta=\cos\) \(\gamma=\frac{1}{\sqrt{3}}\)
Note: Angles are acute. So positive sign of \(\frac{1}{\sqrt{3}}\) is considered.
\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) and given \(\alpha=\beta=\gamma\)
\(\therefore 3 \cos ^{2} \alpha=1 \Rightarrow \cos ^{2} \alpha=\frac{1}{3} \Rightarrow \cos \alpha=\cos \beta=\cos\) \(\gamma=\frac{1}{\sqrt{3}}\)
Note: Angles are acute. So positive sign of \(\frac{1}{\sqrt{3}}\) is considered.
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