Question

Prove that $(\vec a \ + \vec b) . (\vec a + \vec b)=\arrowvert\vec a \arrowvert ^2 + \arrowvert\vec b \arrowvert ^2$, if and only if $\vec a , \vec b$ are perpendicular, given $\vec a \neq 0, \vec b \neq 0$.

Answer 1

Answer:

$\text{\bf\,Concept used:}$

$\vec{a}\text{ and }\vec{b} \text{are perpendicular }\iff \vec{a}.\vec{b}=0$

$\text{Given:}$

$(\vec{a}+\vec{b}).(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$

$\iff\:|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|\vec{a}|^2+|\vec{b}|^2$

$\iff\:2\,\vec{a}.\vec{b}=0$

$\iff\:\vec{a}.\vec{b}=0$

$\iff\:\vec{a}\:\perp\:\vec{b}$

Answer 2

Step-by-step explanation:-

In this question,

We need to prove

$\mathbf{\vec{a}\text{ and }\vec{b} \textrm{are perpendicular }\iff \vec{a}.\vec{b}=0}$

Proof-:

$(\vec{a}+\vec{b}).(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$

On multiplying we get,

$\iff\:|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|\vec{a}|^2+|\vec{b}|^2$

$\iff\:2\,\vec{a}.\vec{b}=0$

$\iff\:\vec{a}.\vec{b}=0$        

using the concept of dot product we get i.e

 {$\vec{a}.\vec{b}Cos90\ =\ 0$}

Hence,$\mathbf{\iff\:\vec{a}\:\perp\:\vec{b}}$