If \( |\vec{a}|=\sqrt{26},|\vec{b}|=7,|\vec{a}\times\vec{b}|=35 \) and \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \) then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| (A) | \( \sin\theta \) | (I) | 61 |
| (B) | \( \cos\theta \) | (II) | \( 5/\sqrt{26} \) |
| (C) | \( \vec{a}\cdot\vec{b} \) | (III) | \( 1/\sqrt{26} \) |
| (D) | \( |\vec{a}-\vec{b}|^2 \) | (IV) | 7 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
The results used are \( |\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin\theta \) and \( \vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta \).
(A) \( \sin\theta \)
This matches (II) \( 5/\sqrt{26} \).
\( \displaystyle \sin\theta=\frac{|\vec{a}\times\vec{b}|}{|\vec{a}|\,|\vec{b}|} \)
\( \displaystyle =\frac{35}{7\sqrt{26}} \)
\( \displaystyle =\frac{5}{\sqrt{26}} \)
(B) \( \cos\theta \)
This matches (III) \( 1/\sqrt{26} \).
\( \displaystyle \cos^2\theta=1-\sin^2\theta \)
\( \displaystyle =1-\frac{25}{26}=\frac{1}{26} \)
The magnitude of the cross product gives only \( \sin\theta \), which is the same for \( \theta \) and \( 180^\circ-\theta \); List-II has only the positive value, so \( \theta \) is acute and \( \cos\theta=\frac{1}{\sqrt{26}} \).
(C) \( \vec{a}\cdot\vec{b} \)
This matches (IV) 7.
\( \displaystyle \vec{a}\cdot\vec{b}=\sqrt{26}\times 7\times\frac{1}{\sqrt{26}}=7 \)
(D) \( |\vec{a}-\vec{b}|^2 \)
This matches (I) 61.
\( \displaystyle |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle =26+49-14=61 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).