Match List-I with List-II
Let \(\theta\) be the angle between the vector \(\vec a\) and \(\vec b\).
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\vec a.\vec b\) | (I) | \(\displaystyle\frac{\vec a.\vec b}{|\vec b|^2}\vec b\) |
| (B) | \(\vec a\times\vec b\) | (II) | \(\vec a.\vec b=0\) |
| (C) | Projection vector of \(\vec a\) on \(\vec b\) (≠0) | (III) | \(|\vec a||\vec b| \sin\theta\,\hat n\) where \(\hat n\) is a unit vector perpendicular to both \(\vec a\) and \(\vec b\) |
| (D) | \(\vec a\) and \(\vec b\) are orthogonal vectors | (IV) | \(|\vec a||\vec b| \cos \theta\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(\vec a.\vec b\)
This matches (IV) \(|\vec a||\vec b|\cos\theta\), the definition of the scalar (dot) product.
(B) \(\vec a\times\vec b\)
This matches (III) \(|\vec a||\vec b|\sin\theta\,\hat n\), the definition of the vector (cross) product.
(C) Projection vector of \(\vec a\) on \(\vec b\)
This matches (I) \(\frac{\vec a.\vec b}{|\vec b|^2}\vec b\). The scalar projection \(\frac{\vec a.\vec b}{|\vec b|}\) times the unit vector \(\frac{\vec b}{|\vec b|}\) gives
\( \displaystyle \frac{\vec a.\vec b}{|\vec b|}\cdot\frac{\vec b}{|\vec b|} = \frac{\vec a.\vec b}{|\vec b|^2}\,\vec b \)
(D) \(\vec a\) and \(\vec b\) are orthogonal vectors
This matches (II) \(\vec a.\vec b = 0\). Orthogonal means \(\theta = \frac{\pi}{2}\), so \(\cos\theta = 0\) and \(\vec a.\vec b = 0\).
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).