Question

25. Two vectors of magnitudes 4 and 6 are acting
through a point. If the
magnitude of the
resultant is R​

Answer 1

Topic :- Vectors

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Two vectors of magnitude 4 & 6 and are acting through a point. If the magnitude of the resultant is R , then

A) 4 < R < 6

B) 4 < R < 10

C) 2 ≤ R ≤ 10

D) 2 ≥ R ≥ 10

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• Two vectors of magnitudes 4 and 6 are acting through a points.

• Let the magnitude of the two vectors be as :

$\footnotesize \bullet \: \displaystyle \:\sf|\overrightarrow{A}|= 6 \: units$

$\footnotesize \bullet \: \displaystyle \:\sf|\overrightarrow{B}|= 4\: units$

$\tiny\dag\:\underline{\sf For \: (\theta) = 0^{\circ} , Parallel \: vectors : }$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB\cos(\theta)}$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB\cos( {0}^{ \circ} )}$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB}$

$\longrightarrow\:\:\sf R =\sqrt{(A+B)^2}$

$\longrightarrow\:\:\sf R =A+B$

$\longrightarrow\:\:\sf R =6+4$

$\longrightarrow\:\: \underline{ \underline{\sf R =10 \: units}}$

• Here,the Resultant R is maximum.

$\tiny\dag\:\underline{\sf For \: (\theta) = 180^{\circ} ,Anti-Parallel \: vectors : }$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB\cos(\theta)}$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB\cos( {180}^{ \circ} )}$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2+2AB \times - 1 }$

$\longrightarrow\:\:\sf R =\sqrt{A^2+B^2 - 2AB }$

$\longrightarrow\:\:\sf R =\sqrt{(A - B)^2 }$

$\longrightarrow\:\:\sf R =A - B$

$\longrightarrow\:\:\sf R =6 - 4$

$\longrightarrow\:\: \underline{\underline{\sf R =2 \: units}}$

• Here,the Resultant R is minimum.

Hence,the magnitude of Resultant for angle 'θ' is :

$\longrightarrow\:\:\sf | A - B | \leqslant R \leqslant A + B$

$\longrightarrow\:\: \underline{ \boxed{\sf 2 \leqslant R \leqslant 10}}$

Hence, option (C) is the correct option.

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