Match List-I with List-II
| List-I | List-II (angle) | ||
| (A) | Angle between \(\hat{\imath}+\hat{\jmath}\) and \(\hat{\jmath}+\hat{k}\) | (I) | \(\displaystyle\frac{\pi}{2}\) |
| (B) | Angle between \(2\hat{\imath}-\hat{\jmath}+\hat{k}\) and \(\hat{\imath}+\hat{\jmath}-\hat{k}\) | (II) | \(\displaystyle\frac{\pi}{4}\) |
| (C) | Angle between \(\vec{a}\) and \(\vec{b}\), where \(|\vec{a}| = |\vec{b}| = 2\) and \(\vec{a}\cdot\vec{b} = -2\sqrt{2}\) | (III) | \(\displaystyle\frac{3\pi}{4}\) |
| (D) | Angle between non-zero vectors \(\vec{a}\) and \(\vec{b}\) with \(|\vec{a}\times\vec{b}| = \vec{a}\cdot\vec{b}\) | (IV) | \(\displaystyle\frac{\pi}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) Angle between \(\hat{\imath}+\hat{\jmath}\) and \(\hat{\jmath}+\hat{k}\)
This matches (IV) \(\frac{\pi}{3}\).
The scalar product is \(0+1+0 = 1\) and each magnitude is \(\sqrt{2}\):
\( \displaystyle \cos\theta = \frac{1}{\sqrt{2}\times\sqrt{2}} = \frac{1}{2} \)
(B) Angle between \(2\hat{\imath}-\hat{\jmath}+\hat{k}\) and \(\hat{\imath}+\hat{\jmath}-\hat{k}\)
This matches (I) \(\frac{\pi}{2}\).
\((2)(1)+(-1)(1)+(1)(-1) = 0\)
A zero scalar product means the vectors are perpendicular.
(C) \(|\vec{a}| = |\vec{b}| = 2\), \(\vec{a}\cdot\vec{b} = -2\sqrt{2}\)
This matches (III) \(\frac{3\pi}{4}\).
\( \displaystyle \cos\theta = \frac{-2\sqrt{2}}{2\times 2} = -\frac{1}{\sqrt{2}} \)
(D) \(|\vec{a}\times\vec{b}| = \vec{a}\cdot\vec{b}\)
This matches (II) \(\frac{\pi}{4}\).
\(|\vec{a}|\,|\vec{b}|\sin\theta = |\vec{a}|\,|\vec{b}|\cos\theta\)
\(\tan\theta = 1\)
Since \(\sin\theta\) and \(\cos\theta\) cannot both be zero, \(\cos\theta \ne 0\), and \(\tan\theta = 1\) with \(0 \le \theta \le \pi\) gives \(\theta = \frac{\pi}{4}\).