If two vectors A→ and B→ having equal magnitude R are inclined at an angle θ, then

Correct answer: (C) A→+B→=2Rcos(θ)/2. The magnitude of resultant vector A→+B→ is given by, R'=√(A^2+B^2+2ABcosθ) Here A=B=R Then R'=√(R^2+R^2+2R^2cosθ) =R√2√(1+cosθ) =√2R√(2cos^2θ2) =2Rcos(θ)/2

Q61 / 90 · JEE Main 2024 (31 Jan Shift 2) · Physics

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $θ$ , then

A
$|\vec{A} - \vec{B}| = \sqrt{2} R \sin (\frac{θ}{2})$
B
$|\vec{A} + \vec{B}| = 2 R \sin (\frac{θ}{2})$
Correct answer
C
$|\vec{A} + \vec{B}| = 2 R \cos (\frac{θ}{2})$
D
$|\vec{A} - \vec{B}| = 2 R \cos (\frac{θ}{2})$

Solution

The magnitude of resultant vector $\vec{A} + \vec{B}$ is given by,

$R' = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$

Here $A = B = R$

Then $R' = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos θ}$

$= R \sqrt{2} \sqrt{1 + \cos θ}$

$= \sqrt{2} R \sqrt{2 \cos^{2} \frac{θ}{2}}$

$= 2 R \cos \frac{θ}{2}$

Hence, the correct answer is option: 'C'.

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