Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Angle between \(\hat{\imath}-\hat{\jmath}\) and \(\hat{\imath}+\hat{\jmath}\) | (I) | \(\pi\) |
| (B) | Angle between \(\hat{\imath}-\hat{\jmath}+\hat{k}\) and \(-\hat{\imath}+\hat{\jmath}-\hat{k}\) | (II) | \(\displaystyle\frac{3\pi}{4}\) |
| (C) | Angle between \(\hat{\imath}+\hat{\jmath}\) and \(-\hat{\imath}\) | (III) | \(\displaystyle\frac{\pi}{4}\) |
| (D) | Angle between \(\hat{\imath}+\hat{k}\) and \(\hat{k}\) | (IV) | \(\displaystyle\frac{\pi}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
The angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) (with \(0 \le \theta \le \pi\)) comes from the dot product:
\( \displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} \)
(A) Angle between \(\hat{\imath}-\hat{\jmath}\) and \(\hat{\imath}+\hat{\jmath}\)
This matches (IV) \(\frac{\pi}{2}\):
\((1)(1) + (-1)(1) = 0\), so the vectors are perpendicular.
(B) Angle between \(\hat{\imath}-\hat{\jmath}+\hat{k}\) and \(-\hat{\imath}+\hat{\jmath}-\hat{k}\)
This matches (I) \(\pi\). The second vector is −1 times the first, so they point in opposite directions:
\( \displaystyle \cos\theta = \frac{-1 - 1 - 1}{\sqrt{3}\,\sqrt{3}} = -1 \)
(C) Angle between \(\hat{\imath}+\hat{\jmath}\) and \(-\hat{\imath}\)
This matches (II) \(\frac{3\pi}{4}\):
\( \displaystyle \cos\theta = \frac{(1)(-1) + 0}{\sqrt{2} \times 1} = -\frac{1}{\sqrt{2}} \)
(D) Angle between \(\hat{\imath}+\hat{k}\) and \(\hat{k}\)
This matches (III) \(\frac{\pi}{4}\):
\( \displaystyle \cos\theta = \frac{0 + 0 + 1}{\sqrt{2} \times 1} = \frac{1}{\sqrt{2}} \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).