Question

The angle between the diagonals of the parallelogram whose adjacent sides are \(2i + 4j - 5k\) and \(i + 2j + 3k\) is

• A. \( \cos^{-1} \left(\frac{7}{69}\right) \)
• B. \( \cos^{-1} \left(\frac{1}{\sqrt{69}}\right) \)
• C. \( \cos^{-1} \left(\frac{1}{7}\right) \)
• D. \( \cos^{-1} \left(\frac{31}{7\sqrt{69}}\right) \)

Hint:

Remember that the cosine of the angle between two vectors can be calculated using the dot product formula. The angle between diagonals of a parallelogram can reveal symmetrical properties and geometrical insights into its shape.

Answer 1

The Correct Option is D

Step 1: Calculate the diagonals of the parallelogram. The diagonals are given by the vectors: \[ \mathbf{d}_1 = \mathbf{u} + \mathbf{v} = (2i + 4j - 5k) + (i + 2j + 3k) = 3i + 6j - 2k, \] \[ \mathbf{d}_2 = \mathbf{u} - \mathbf{v} = (2i + 4j - 5k) - (i + 2j + 3k) = i + 2j - 8k. \] Step 2: Use the dot product to find the cosine of the angle \(\theta\) between the diagonals: \[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (3i + 6j - 2k) \cdot (i + 2j - 8k) = 3 + 12 + 16 = 31, \] \[ \|\mathbf{d}_1\| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{49} = 7, \] \[ \|\mathbf{d}_2\| = \sqrt{1^2 + 2^2 + (-8)^2} = \sqrt{69}. \] \[ \cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{\|\mathbf{d}_1\| \|\mathbf{d}_2\|} = \frac{31}{7\sqrt{69}}. \] Step 3: Compute the angle \(\theta\) using the cosine inverse function: \[ \theta = \cos^{-1} \left(\frac{31}{7\sqrt{69}}\right). \]