Question
Prove that sum and differences of two equal and perpendicular vectors will also be equal and perpendicular.
✓
Answer
$\begin{aligned}
R_1 & =\sqrt{A^2+A^2}=\sqrt{2} A \\
\tan \theta & =\frac{A}{A}=1 \quad \therefore \quad \theta=45^{\circ}
\end{aligned}$
Magnitude of $R _2$ :
$R_2=\sqrt{A^2+A^2}=\sqrt{2} A$
If $\vec{R}_2$ makes angle $\theta$ with vector $\vec{A}$ then,
$\tan \theta=\frac{A}{A}=1$
$\therefore \quad \theta=45^{\circ}$
It is clear that $\vec{R}_1$ and $\vec{R}_2$ are equal and perpendicular to each other.
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