Question

If \(\alpha\) is the angle between any two diagonals of a cube and \(\beta\) is the angle between a diagonal of a cube and a diagonal of its face, which intersects this diagonal of the cube then \( \cos\alpha + \cos^2\beta = \)

• A. \( \frac{5}{9} \)
• B. \( \frac{2}{9} \)
• C. 1
• D. \( \frac{2}{3} \)

Hint:

For a cube, you can memorize these standard angles: - Angle between two body diagonals: \( \cos^{-1}(1/3) \). - Angle between a body diagonal and a face diagonal: \( \cos^{-1}(\sqrt{2/3}) \). - Angle between a body diagonal and an edge: \( \cos^{-1}(1/\sqrt{3}) \).

Answer 1

The Correct Option is C

Let the vertices of a cube of side length 'a' be placed at the origin (0,0,0) and along the positive axes.
Part 1: Find \( \cos\alpha \), the cosine of the angle between two body diagonals.
Let's take the diagonal from the origin (0,0,0) to (a,a,a), represented by vector \( \vec{d_1} = a\vec{i}+a\vec{j}+a\vec{k} \).
And the diagonal from (a,0,0) to (0,a,a), represented by vector \( \vec{d_2} = (0-a)\vec{i}+(a-0)\vec{j}+(a-0)\vec{k} = -a\vec{i}+a\vec{j}+a\vec{k} \).
\( \cos\alpha = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|} \).
\( \vec{d_1} \cdot \vec{d_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2+a^2+a^2=a^2 \).
\( |\vec{d_1}| = \sqrt{a^2+a^2+a^2} = a\sqrt{3} \). \( |\vec{d_2}| = \sqrt{(-a)^2+a^2+a^2} = a\sqrt{3} \).
\( \cos\alpha = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3} \).
Part 2: Find \( \cos\beta \), the cosine of the angle between a body diagonal and a face diagonal.
Let's use the same body diagonal \( \vec{d_1} = a\vec{i}+a\vec{j}+a\vec{k} \).
An intersecting face diagonal can be taken from the origin (0,0,0) to (a,a,0), represented by vector \( \vec{f} = a\vec{i}+a\vec{j} \).
\( \cos\beta = \frac{\vec{d_1} \cdot \vec{f}}{|\vec{d_1}||\vec{f}|} \).
\( \vec{d_1} \cdot \vec{f} = (a)(a) + (a)(a) + (a)(0) = 2a^2 \).
\( |\vec{f}| = \sqrt{a^2+a^2+0^2} = a\sqrt{2} \).
\( \cos\beta = \frac{2a^2}{(a\sqrt{3})(a\sqrt{2})} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}} \).
Part 3: Calculate the required expression.
We need to find \( \cos\alpha + \cos^2\beta \).
\( \cos^2\beta = \left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3} \).
\( \cos\alpha + \cos^2\beta = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1 \).