Question

Let \(OA\), \(OB\), \(OC\) lying along the \(X\)-, \(Y\)- and \(Z\)-axes respectively represent the coterminous edges of a rectangular parallelepiped. If \(OA=1\), \(OB=2\), \(OC=3\), then the angle between a pair of diagonals of the parallelepiped drawn through the vertices \(O\) and \(A\) is

• A. \(\dfrac{\pi}{3}\)
• B. \(\cos^{-1}\left(\dfrac{5}{7}\right)\)
• C. \(\cos^{-1}\left(\dfrac{6}{7}\right)\)
• D. \(\dfrac{\pi}{4}\)

Hint:

For questions involving angles between diagonals of a cuboid, first express the diagonals as vectors. The dot product formula \[ \cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \] usually leads directly to the answer with minimal computation.

Answer 1

The Correct Option is C

Concept: A rectangular parallelepiped (cuboid) can be represented using vectors. The angle between two diagonals can be determined by finding the direction vectors of those diagonals and then using the dot product formula. If vectors \(\mathbf{u}\) and \(\mathbf{v}\) make an angle \(\theta\), then \[ \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}} {|\mathbf{u}|\,|\mathbf{v}|}. \] Therefore, our strategy will be:
• Assign coordinates to all vertices.
• Determine the diagonals passing through \(O\) and \(A\).
• Find their direction vectors.
• Use the dot product formula.

Step 1: Assign coordinates. Since \[ OA=1,\qquad OB=2,\qquad OC=3, \] take \[ O=(0,0,0), \] \[ A=(1,0,0), \] \[ B=(0,2,0), \] \[ C=(0,0,3). \] The opposite vertex of the cuboid is \[ P=(1,2,3). \]

Step 2: Identify the diagonals through \(O\) and \(A\). One space diagonal through \(O\) is \[ OP. \] Its direction vector is \[ \mathbf{u}=(1,2,3). \] The space diagonal through \(A\) joins \[ A=(1,0,0) \] to \[ Q=(0,2,3). \] Its direction vector is \[ \mathbf{v}=(-1,2,3). \]

Step 3: Compute the dot product. \[ \mathbf{u}\cdot\mathbf{v} = (1)(-1)+(2)(2)+(3)(3). \] \[ =-1+4+9 =12. \]

Step 4: Find the magnitudes. \[ |\mathbf{u}| = \sqrt{1^2+2^2+3^2} = \sqrt{14}. \] Similarly, \[ |\mathbf{v}| = \sqrt{(-1)^2+2^2+3^2} = \sqrt{14}. \]

Step 5: Find the angle. Using the dot product formula, \[ \cos\theta = \frac{12}{\sqrt{14}\sqrt{14}} = \frac{12}{14} = \frac{6}{7}. \] Therefore, \[ \theta = \cos^{-1}\left(\frac{6}{7}\right). \] Hence, \[ \boxed{\theta=\cos^{-1}\left(\frac{6}{7}\right)}. \]