Question
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular vectors, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=3$ and $|\vec{\text{a}}|=5,$ find the value of $\big|\vec{\text{b}}\big|.$
✓
Answer
Disclamer: $\big|\vec{\text{a}}+\vec{\text{b}}\big|=3$ has been taken in order to solve question.
It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular vectors.
$\therefore\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=13$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=169$
$\Rightarrow|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}+\big|\vec{\text{b}}\big|^2=169$
$\Rightarrow25+2\times0+\big|\vec{\text{b}}\big|^2=169$ [using (1)]
$\Rightarrow\big|\vec{\text{b}}\big|^2=169-25=144$
$\Rightarrow\big|\vec{\text{b}}\big|=12$
Thus, the value of $\big|\vec{\text{b}}\big|$ is 12.
It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular vectors.
$\therefore\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=13$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=169$
$\Rightarrow|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}+\big|\vec{\text{b}}\big|^2=169$
$\Rightarrow25+2\times0+\big|\vec{\text{b}}\big|^2=169$ [using (1)]
$\Rightarrow\big|\vec{\text{b}}\big|^2=169-25=144$
$\Rightarrow\big|\vec{\text{b}}\big|=12$
Thus, the value of $\big|\vec{\text{b}}\big|$ is 12.
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