The angle between vector (A) and (A - B) is: (a) tan ⁻¹((-B/2)/(A - B√(3)/2)) (b) tan ⁻¹(A/0.7B) (c) tan ⁻¹((√(3)B)/(2A - B)) (d) tan ⁻¹((Bcos θ)/(A - Bsin θ))

Answer: (c). AB-BA - Bβ60° From the figure the angle between A and B (tail to tail) is 180° - 120° = 60°. Take A along x: A - B = (A - B/2)i - (√(3)B)/(2)j. tan β = (√(3)B/2)/(A - B/2) = (√(3)B)/(2A - B) So the answer is (c) tan ⁻¹((√(3)B)/(2A - B)).

The angle between vector (A) and (A - B) is: | JEE Main 2021 (26 Aug, Shift 2)

Q20JEE Main 2021 · 26 AUG · SHIFT 2Medium

Addition & Subtraction of Vectors

The angle between vector $\left(\vec{A}\right)$ and $\left(\vec{A} - \vec{B}\right)$ is:

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Answer: (c)

From the figure the angle between $\vec{A}$ and $\vec{B}$ (tail to tail) is $180^\circ - 120^\circ = 60^\circ$ .
Take $\vec{A}$ along $x$ : $\vec{A} - \vec{B} = \left(A - \frac{B}{2}\right)\hat{i} - \frac{\sqrt{3}B}{2}\hat{j}$ .
$\tan\beta = \frac{\sqrt{3}B/2}{A - B/2} = \frac{\sqrt{3}B}{2A - B}$
So the answer is (c) $\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)$ .

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Addition & Subtraction of Vectors question from JEE Main 2021 (26 Aug, Shift 2)

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