First slide

Scalars and Vectors

Question

The resultant of $\vec{A} + \vec{B}$ is ${\vec{R}}_{1}$ . On reversing the vector $\vec{B}$ the resultant becomes ${\vec{R}}_{2}$ . What is the value of $R_{1}^{2} + R_{2}^{2}$ ?

Moderate

A

$A^{2} + B^{2}$
B

$A^{2} - B^{2}$
C

$2 (A^{2} + B^{2})$
D

$2 (A^{2} - B^{2})$

Solution

$\begin{array}{l} {\vec{R}}_{1} = \vec{A} + \vec{B}, {\vec{R}}_{2} = \vec{A} - \vec{B} \\ ∴ R_{1}^{2} + R_{2}^{2} = {(\sqrt{A^{2} + B^{2} + 2 A B \cos θ})}^{2} + {(\sqrt{A^{2} + B^{2} - 2 A B \cos θ})}^{2} = 2 (A^{2} + B^{2}) \end{array}$

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