The adjacent sides of a parallelogram are given by the vectors \(\vec{a}=2\hat{\imath}-\hat{\jmath}-\hat{k}\) and \(\vec{b}=-\hat{\jmath}+\hat{k}\). The acute angle between the diagonals of the parallelogram is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{3}\)
The diagonals of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) are \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\):
\(\vec{d}_1 = \vec{a}+\vec{b} = 2\hat{\imath}-2\hat{\jmath}\)
\(\vec{d}_2 = \vec{a}-\vec{b} = 2\hat{\imath}-2\hat{k}\)
\(\vec{d}_1\cdot\vec{d}_2 = (2)(2)+(-2)(0)+(0)(-2) = 4\)
\(|\vec{d}_1| = \sqrt{8}, \quad |\vec{d}_2| = \sqrt{8}\)
Let \(\theta\) be the angle between the diagonals:
\( \displaystyle \cos\theta = \frac{4}{\sqrt{8}\times\sqrt{8}} = \frac{1}{2} \)
\( \displaystyle \theta = \frac{\pi}{3} \)
Since \(\cos\theta \gt 0\), this angle is already the acute angle.
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{6}\) — Incorrect. \(\cos\displaystyle\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), but \(\cos\theta = \displaystyle\frac{1}{2}\).
Option (2) \(\displaystyle\frac{\pi}{4}\) — Incorrect. \(\cos\displaystyle\frac{\pi}{4} = \frac{1}{\sqrt{2}}\), but \(\cos\theta = \displaystyle\frac{1}{2}\).
Option (3) \(\displaystyle\frac{\pi}{2}\) — Incorrect. The diagonals are perpendicular only when \(|\vec{a}| = |\vec{b}|\) (a rhombus); here \(|\vec{a}| = \sqrt{6}\) and \(|\vec{b}| = \sqrt{2}\), and \(\vec{d}_1\cdot\vec{d}_2 = 4 \ne 0\).