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If two vectors $\mathbf{A}$ and $\mathbf{B}$ are mutually perpendicular, then the component of $\mathbf{A}-\mathbf{B}$ along the direction of $\mathbf{A}+\mathbf{B}$ is
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The correct answer is:
$\frac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$
The two vectors $\mathbf{A}$ and $\mathbf{B}$ are mutually perpendicular to each other
A. $\mathbf{B}=0$
$\Rightarrow \quad A B \cos \theta=0$
$\Rightarrow \quad \cos \theta=0 \Rightarrow \theta=90^{\circ}$
$|\mathbf{A}-\mathbf{B}|$
$\begin{aligned} & =\sqrt{A^2+B^2-2 A B \cos 90^{\circ}} \\ & =\sqrt{A^2+B^2}\end{aligned}$
Unit vector along $\mathbf{A}-\mathbf{B}$
$=\frac{\mathbf{A}-\mathbf{B}}{|\mathbf{A}-\mathbf{B}|}=\frac{\mathbf{A}-\mathbf{B}}{\sqrt{A^2+B^2}}$
component of $(\mathbf{A}+\mathbf{B})$ in direction of $(\mathbf{A}-\mathbf{B})$ will be given as,
$=\frac{(\mathbf{A}+\mathbf{B}) \cdot(\mathbf{A}-\mathbf{B})}{|\mathbf{A}-\mathbf{B}|}$
$=\frac{\mathbf{A} \cdot \mathbf{A}-\mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{A}-\mathbf{B} \cdot \mathbf{B}}{\sqrt{A^2+B^2}}$
$=\frac{A^2-B^2}{\sqrt{A^2+B^2}}=\frac{|\mathbf{A}|^2-|\mathbf{B}|^2}{\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2}}$
A. $\mathbf{B}=0$
$\Rightarrow \quad A B \cos \theta=0$
$\Rightarrow \quad \cos \theta=0 \Rightarrow \theta=90^{\circ}$
$|\mathbf{A}-\mathbf{B}|$
$\begin{aligned} & =\sqrt{A^2+B^2-2 A B \cos 90^{\circ}} \\ & =\sqrt{A^2+B^2}\end{aligned}$
Unit vector along $\mathbf{A}-\mathbf{B}$
$=\frac{\mathbf{A}-\mathbf{B}}{|\mathbf{A}-\mathbf{B}|}=\frac{\mathbf{A}-\mathbf{B}}{\sqrt{A^2+B^2}}$
component of $(\mathbf{A}+\mathbf{B})$ in direction of $(\mathbf{A}-\mathbf{B})$ will be given as,
$=\frac{(\mathbf{A}+\mathbf{B}) \cdot(\mathbf{A}-\mathbf{B})}{|\mathbf{A}-\mathbf{B}|}$
$=\frac{\mathbf{A} \cdot \mathbf{A}-\mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{A}-\mathbf{B} \cdot \mathbf{B}}{\sqrt{A^2+B^2}}$
$=\frac{A^2-B^2}{\sqrt{A^2+B^2}}=\frac{|\mathbf{A}|^2-|\mathbf{B}|^2}{\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2}}$
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