Question

If each side of a cube is x, then the angle between the diagonals of the cube is :

• A. \(\cos^{-1}(\frac{1}{\sqrt3})\)
• B. \(\cos^{-1}(\frac{1}{3})\)<
• C. \(\cos^{-1}(-\frac{1}{\sqrt3})\)
• D. \(\cos^{-1}(-\frac{1}{3})\)

Answer 1

The Correct Option is B

The problem involves finding the angle between the diagonals of a cube where each side has a length of \( x \). To solve this, we consider a cube with vertices at \((0,0,0)\), \((x,0,0)\), \((0,x,0)\), \((0,0,x)\), \((x,x,0)\), \((0,x,x)\), \((x,0,x)\), and \((x,x,x)\). The main body diagonals of the cube run from \((0,0,0)\) to \((x,x,x)\) and from \((x,0,0)\) to \((0,x,x)\).
Let's calculate the vectors representing these diagonals. The first diagonal vector is \(\vec{a} = x\hat{i} + x\hat{j} + x\hat{k}\) and the second diagonal vector is \(\vec{b} = -x\hat{i} + x\hat{j} + x\hat{k}\).
Next, we find the dot product \(\vec{a} \cdot \vec{b}\):
\( \vec{a} \cdot \vec{b} = (x)(-x) + (x)(x) + (x)(x) = -x^2 + x^2 + x^2 = x^2 \).
The magnitudes of \(\vec{a}\) and \(\vec{b}\) are:
\(|\vec{a}| = |\vec{b}| = \sqrt{x^2 + x^2 + x^2} = \sqrt{3x^2} = x\sqrt{3}\).
The cosine of the angle \(\theta\) between these vectors is given by:
\(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{x^2}{x\sqrt{3} \cdot x\sqrt{3}} = \frac{x^2}{3x^2} = \frac{1}{3}\).
Therefore, the angle \(\theta\) is \(\cos^{-1}\left(\frac{1}{3}\right)\).
The correct answer is \(\cos^{-1}\left(\frac{1}{3}\right)\).