Let \(\vec{a}\) and \(\vec{b}\) be non-zero vectors. Match List-I with List-II
| List-I (condition) | List-II (angle between \(\vec{a}\) and \(\vec{b}\)) | ||
| (A) | \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\) | (I) | \(\displaystyle\frac{\pi}{3}\) |
| (B) | \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\) | (II) | \(\pi\) |
| (C) | \(|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{a}|\,|\vec{b}|\) | (III) | 0 |
| (D) | \(\vec{a}\cdot\vec{b} = -|\vec{a}|\,|\vec{b}|\) | (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)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each condition gives \(\cos\theta\) through \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\), with \(0 \le \theta \le \pi\).
(A) \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\)
This matches (III) 0.
\(\cos\theta = 1\)
The vectors point the same way.
(B) \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\)
This matches (IV) \(\frac{\pi}{2}\).
\(|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} = |\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b}\)
\(4\,\vec{a}\cdot\vec{b} = 0\)
A zero scalar product of non-zero vectors means they are perpendicular.
(C) \(|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{a}|\,|\vec{b}|\)
This matches (I) \(\frac{\pi}{3}\).
\(|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} = |\vec{a}|^2+|\vec{b}|^2+|\vec{a}|\,|\vec{b}|\)
\(2|\vec{a}|\,|\vec{b}|\cos\theta = |\vec{a}|\,|\vec{b}|\)
\( \displaystyle \cos\theta = \frac{1}{2} \)
(D) \(\vec{a}\cdot\vec{b} = -|\vec{a}|\,|\vec{b}|\)
This matches (II) \(\pi\).
\(\cos\theta = -1\)
The vectors point in opposite directions.