Match List I with List II
| LIST I | LIST II | ||
| A. | For any vector \(\vec{a}\), \(\vec{a}\times\vec{a}\) | I. | \(\vec{a}^2\) |
| B. | For any vector \(\vec{a}\), \(\vec{a}\cdot\vec{a}\) | II. | 0 |
| C. | If either \(\vec{a}=\vec{0}\) or \(\vec{b}=\vec{0}\), then \(\vec{a}\cdot\vec{b}\) is | III. | \(|\vec{a}||\vec{b}|\) |
| D. | If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta=0\), then \(\vec{a}\cdot\vec{b}\) is | IV. | \(\vec{0}\) |
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 cross product of two vectors is a vector; the dot product is a scalar, \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\).
(A) For any vector \(\vec{a}\), \(\vec{a}\times\vec{a}\)
This matches (IV) \(\vec{0}\). The angle between \(\vec{a}\) and itself is 0, and \(\sin 0 = 0\), so \(\vec{a}\times\vec{a}\) is the zero vector.
(B) For any vector \(\vec{a}\), \(\vec{a}\cdot\vec{a}\)
This matches (I) \(\vec{a}^2\):
\(\vec{a}\cdot\vec{a} = |\vec{a}||\vec{a}|\cos 0 = |\vec{a}|^2\), written \(\vec{a}^2\)
(C) If either \(\vec{a}=\vec{0}\) or \(\vec{b}=\vec{0}\), then \(\vec{a}\cdot\vec{b}\) is
This matches (II) 0. One of \(|\vec{a}|\), \(|\vec{b}|\) is 0, and the dot product is the scalar 0 (not the zero vector).
(D) If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta=0\), then \(\vec{a}\cdot\vec{b}\) is
This matches (III) \(|\vec{a}||\vec{b}|\):
\(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos 0 = |\vec{a}||\vec{b}|\)