If two vectors A and B are mutually perpendic | vedclass

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(C) Given that vectors $A$ and $B$ are mutually perpendicular,so $A \cdot B = 0$.We need to find the component of vector $(A-B)$ along the direction o

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$\frac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$

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(C) Given that vectors $A$ and $B$ are mutually perpendicular,so $A \cdot B = 0$.We need to find the component of vector $(A-B)$ along the direction of vector $(A+B)$.The component of a vector $P$ along the direction of vector $Q$ is given by the formula: $	ext{Component} = \frac{P \cdot Q}{|Q|}$.Here,$P = A-B$ and $Q = A+B$.First,calculate the magnitude of $(A+B)$:$|A+B| = \sqrt{(A+B) \cdot (A+B)} = \sqrt{A \cdot A + B \cdot B + 2(A \cdot B)} = \sqrt{|A|^2 + |B|^2 + 0} = \sqrt{|A|^2 + |B|^2}$.Now,calculate the dot product $(A-B) \cdot (A+B)$:$(A-B) \cdot (A+B) = A \cdot A + A \cdot B - B \cdot A - B \cdot B = |A|^2 + 0 - 0 - |B|^2 = |A|^2 - |B|^2$.Finally,the component is:$	ext{Component} = \frac{(A-B) \cdot (A+B)}{|A+B|} = \frac{|A|^2 - |B|^2}{\sqrt{|A|^2 + |B|^2}}$.

If two vectors $A$ and $B$ are mutually perpendicular,then the component of $A-B$ along the direction of $A+B$ is

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