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If $\mathbf{A}+\mathbf{B}=\mathbf{C}$ and that $\mathbf{C}$ is perpendicular to $\mathbf{A}$. What is the angle between $\mathbf{A}$ and $\mathbf{B}$, if $|\mathbf{A}|=|\mathbf{C}|$ ?
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Verified Answer
The correct answer is:
$\frac{3 \pi}{4} \mathrm{rad}$
Given, $A+B=C$
and $\mathrm{B}=\mathrm{C}-\mathrm{A}$
Since, $\mathbf{C} \perp \mathbf{A}$, therefore
$B^2=C^2+A^2$
Also, $|\mathbf{C}|=|\mathbf{A}|$, therefore
$B^2=2 A^2 \Rightarrow B=\sqrt{2} A$
Now, $A^2+B^2+2 A B \cos \theta=C^2=A^2$
$\therefore \quad \cos \theta=-\frac{1}{\sqrt{2}}$
This gives $\quad \theta=\frac{3 \pi}{4} \mathrm{rad}$
and $\mathrm{B}=\mathrm{C}-\mathrm{A}$
Since, $\mathbf{C} \perp \mathbf{A}$, therefore
$B^2=C^2+A^2$
Also, $|\mathbf{C}|=|\mathbf{A}|$, therefore
$B^2=2 A^2 \Rightarrow B=\sqrt{2} A$
Now, $A^2+B^2+2 A B \cos \theta=C^2=A^2$
$\therefore \quad \cos \theta=-\frac{1}{\sqrt{2}}$
This gives $\quad \theta=\frac{3 \pi}{4} \mathrm{rad}$
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