Question:
The angle between a diagonal and one of its edges of a cube is ...................
The angle between a diagonal and one of its edges of a cube is ...................
Show Hint
Take the cube with edge 1, diagonal \((1,1,1)\) and edge \((1,0,0)\).
Updated On: Oct 1, 2026
- \(tan^{-1}(\frac{1}{\sqrt{2}})\)
- \(cos^{-1}(\frac{1}{3})\)
- \(cos^{-1}(\frac{1}{2\sqrt{2}})\)
- \(tan^{-1}(\sqrt{2})\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Place the cube with one vertex at the origin. A body diagonal has direction ratios \(1,1,1\), and an edge from the same vertex has direction \(1,0,0\).
Step 2: Key Formula or Approach
\[ \cos\theta=\frac{1\cdot1+1\cdot0+1\cdot0}{\sqrt3\cdot1}=\frac{1}{\sqrt3} \]
Step 3: Detailed Explanation
Then \(\sin\theta=\sqrt{1-\tfrac13}=\sqrt{\tfrac23}\).
\[ \tan\theta=\frac{\sqrt{2/3}}{1/\sqrt3}=\sqrt2 \]
So \(\theta=\tan^{-1}\sqrt2\), equivalent to \(\cos^{-1}\dfrac{1}{\sqrt3}\).
Final Answer:
The angle is \(\tan^{-1}\sqrt2\), option (D). \[ \boxed{\tan^{-1}\sqrt2\ \text{(D)}} \]
Place the cube with one vertex at the origin. A body diagonal has direction ratios \(1,1,1\), and an edge from the same vertex has direction \(1,0,0\).
Step 2: Key Formula or Approach
\[ \cos\theta=\frac{1\cdot1+1\cdot0+1\cdot0}{\sqrt3\cdot1}=\frac{1}{\sqrt3} \]
Step 3: Detailed Explanation
Then \(\sin\theta=\sqrt{1-\tfrac13}=\sqrt{\tfrac23}\).
\[ \tan\theta=\frac{\sqrt{2/3}}{1/\sqrt3}=\sqrt2 \]
So \(\theta=\tan^{-1}\sqrt2\), equivalent to \(\cos^{-1}\dfrac{1}{\sqrt3}\).
Final Answer:
The angle is \(\tan^{-1}\sqrt2\), option (D). \[ \boxed{\tan^{-1}\sqrt2\ \text{(D)}} \]
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