Question

Angle between a diagonal of a cube and a diagonal of its face which are coterminus is

• A. \( \dfrac{\pi}{2} \)
• B. \( \cos^{-1}\left( \dfrac{2}{\sqrt{3}} \right) \)
• C. \( \cos^{-1}\left( \dfrac{1}{\sqrt{3}} \right) \)
• D. \( \cos^{-1}\left( \dfrac{\sqrt{3}}{2} \right) \)

Hint:

To find the angle between a body diagonal and a face diagonal of a cube, treat them as vectors and use the dot product formula with known coordinates.

Answer 1

The Correct Option is B

Step 1: Understand the geometry A cube has: Face diagonal lying in a square face. Space diagonal (body diagonal) connecting two opposite vertices of the cube. Let the cube be of unit length (side = 1 unit). Take the origin as one vertex and cube aligned along the coordinate axes. 
Step 2: Define vectors Let the origin \( O = (0, 0, 0) \). Consider: Face diagonal: along the bottom face from \( O = (0, 0, 0) \) to \( A = (1, 1, 0) \), so vector \( \vec{v}_1 = \langle 1, 1, 0 \rangle \) Space diagonal (body diagonal): from \( O = (0, 0, 0) \) to \( B = (1, 1, 1) \), so vector \( \vec{v}_2 = \langle 1, 1, 1 \rangle \) 
Step 3: Use dot product to find angle Angle \( \theta \) between two vectors is given by: \[ \cos \theta = \dfrac{\vec{v}_1 \cdot \vec{v}_2}{\|\vec{v}_1\| \cdot \|\vec{v}_2\|} \] Compute: \[ \vec{v}_1 \cdot \vec{v}_2 = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 = 2 \] \[ \|\vec{v}_1\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}, \|\vec{v}_2\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \] Step 4: Compute cosine of the angle \[ \cos \theta = \dfrac{2}{\sqrt{2} \cdot \sqrt{3}} = \dfrac{2}{\sqrt{6}} = \dfrac{2\sqrt{6}}{6} = \dfrac{\sqrt{6}}{3} \] Step 5: Final Answer \[ \theta = \cos^{-1}\left( \dfrac{\sqrt{6}}{3} \right) = \cos^{-1} \left( \dfrac{2}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \right) = \cos^{-1} \left( \dfrac{2}{\sqrt{3}} \right) \]