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$$ \begin{aligned} & \vec{A}+\vec{B}=\vec{C} \quad \vec{A} \\ & C^2=A^2+B^2+2 A B \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $$

and $\tan \alpha=\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}$

When $B$ is doubled, resultant is perpendicular to $\vec{A}$

$$\therefore C_1^2=A^2+4 B^2+4 A B \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$

MHT CET 2025 19th April Evening Shift Physics - Vector Algebra Question 27 English Explanation 2

From right angled triangle PSR

$$ \begin{aligned} & (2 B)^2=C_1^2+A^2 \\ & C_1^2=4 B^2-A^2 \end{aligned} $$

Substituting in (ii) and solving,

$$ \mathrm{A}^2+2 \mathrm{AB} \cos \theta=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$

Substituting (iii) in (i), $\mathrm{C}=\mathrm{B}$

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