Let \(\theta\) be the angle between two vectors \(\vec{a}\) and \(\vec{b}\). Then match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\sin\theta\) | (I) | \(\displaystyle\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\) |
| (B) | \(\cos\theta\) | (II) | \(|\vec{a}\times\vec{b}|\) |
| (C) | Area of the parallelogram with adjacent sides represented by \(\vec{a}\) and \(\vec{b}\) | (III) | \(\displaystyle\frac{\vec{a}.\vec{b}}{|\vec{a}|}\) |
| (D) | Projection of \(\vec{a}\) on \(\vec{b}\) | (IV) | \(\displaystyle\frac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|}\) |
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 dot and cross products give
\(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\)
\(|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)
(A) \(\sin\theta\)
This matches (IV):
\( \displaystyle \sin\theta = \frac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|} \)
(B) \(\cos\theta\)
This matches (I):
\( \displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} \)
(C) Area of the parallelogram with adjacent sides represented by \(\vec{a}\) and \(\vec{b}\)
This matches (II) \(|\vec{a}\times\vec{b}|\). Taking \(|\vec{a}|\) as the base, the height is \(|\vec{b}|\sin\theta\):
\(\text{Area} = |\vec{a}||\vec{b}|\sin\theta = |\vec{a}\times\vec{b}|\)
(D) Projection of \(\vec{a}\) on \(\vec{b}\)
This matches (III), by elimination. The projection of \(\vec{a}\) on \(\vec{b}\) is
\( \displaystyle |\vec{a}|\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).
Note: It seems that there is a mistake in the original NTA paper, which prints (III) as \(\frac{\vec{a}.\vec{b}}{|\vec{a}|}\) when it should read \(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\). The printed entry is the projection of \(\vec{b}\) on \(\vec{a}\), not of \(\vec{a}\) on \(\vec{b}\). The question can still be solved because (A), (B) and (C) match only (IV), (I) and (II), which leaves (III) for (D), and Option (1) is the only option with (A)-(IV), (B)-(I), (C)-(II).