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If $\mathbf{a}$ and $\mathbf{b}$ are vectors such that $|\mathbf{a}+\mathbf{b}|$ $=\mid \mathbf{a}-\mathbf{b}$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is
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Verified Answer
The correct answer is:
$90^{\circ}$
We have, $\quad|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|$
On squaring both sides, we get
$\quad|\mathbf{a}+\mathbf{b}|^{2}=|\mathbf{a}-\mathbf{b}|^{2}$ $\Rightarrow|\mathbf{a}|^{2}+\left.\left|\mathbf{b}^{2}+2 \mathbf{a} \cdot \mathbf{b}=\right| \mathbf{a}\right|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$ $\Rightarrow \quad 4 \mathbf{a} \cdot \mathbf{b}=0$ $\Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0$ $\Rightarrow \mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other. So, angle between them is $90^{\circ} .$ Alternative $\because \quad|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|$ $\therefore \mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other So, angle between $\mathbf{a}$ and $\mathbf{b}$ is $90^{\circ} .$
So, angle between them is $90^{\circ}$.
Alternative
$\therefore$ a and $\mathbf{b}$ are perpendicular to each other So, angle between $\mathbf{a}$ and $\mathbf{b}$ is $90^{\circ}$.
On squaring both sides, we get
$\quad|\mathbf{a}+\mathbf{b}|^{2}=|\mathbf{a}-\mathbf{b}|^{2}$ $\Rightarrow|\mathbf{a}|^{2}+\left.\left|\mathbf{b}^{2}+2 \mathbf{a} \cdot \mathbf{b}=\right| \mathbf{a}\right|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$ $\Rightarrow \quad 4 \mathbf{a} \cdot \mathbf{b}=0$ $\Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0$ $\Rightarrow \mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other. So, angle between them is $90^{\circ} .$ Alternative $\because \quad|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|$ $\therefore \mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other So, angle between $\mathbf{a}$ and $\mathbf{b}$ is $90^{\circ} .$
So, angle between them is $90^{\circ}$.
Alternative
$\therefore$ a and $\mathbf{b}$ are perpendicular to each other So, angle between $\mathbf{a}$ and $\mathbf{b}$ is $90^{\circ}$.
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