# How do we prove the projection law using vectors?

Apr 23, 2018

Let $\vec{a} \mathmr{and} \vec{b}$ are two vectors inclined at an angle $\theta$.

So

$\vec{a} \cdot \vec{b} = \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \theta$

Now vector projection of $\vec{a}$ on to $\vec{b}$ will be
$= \left\mid \vec{a} \right\mid \cos \theta \cdot \text{unit vector of } \vec{b}$

$= \left\mid \vec{a} \right\mid \cos \theta \cdot \frac{\vec{b}}{\left\mid \vec{b} \right\mid}$
$= \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \theta \cdot \frac{\vec{b}}{{\left\mid \vec{b} \right\mid}^{2}}$

$= \left(\vec{a} \cdot \vec{b}\right) \frac{\vec{b}}{{\left\mid \vec{b} \right\mid}^{2}}$

Apr 23, 2018

To prove for $A B C$
$a = b \cos \left(C\right) + c \cos \left(B\right)$ by using vector law.

Proof

Let us consider that three vectors $\vec{a} , \vec{b} \mathmr{and} \vec{c}$ are represented respectively in order by three sides $B C , C A \mathmr{and} A B$ of a $\Delta A B C$.

This means

$\vec{B C} = \vec{a}$

$\vec{C A} = \vec{b}$

$\vec{A B} = \vec{c}$

So we can write

$\vec{B C} + \vec{C A} + \vec{A B} = 0$

$\implies \vec{a} + \vec{b} + \vec{c} = 0$

$\implies \vec{a} \cdot \left(\vec{a} + \vec{b} + \vec{c}\right) = 0$

$\implies \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$

$\implies \left\mid \vec{a} \right\mid \left\mid \vec{a} \right\mid \cos {0}^{\circ} + \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \left(\pi - C\right) + \left\mid \vec{a} \right\mid \left\mid \vec{c} \right\mid \cos \left(\pi - B\right) = 0$

$\implies \left\mid \vec{a} \right\mid \left\mid \vec{a} \right\mid \cdot 1 - \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \left(C\right) - \left\mid \vec{a} \right\mid \left\mid \vec{c} \right\mid \cos \left(B\right) = 0$

$\implies \left\mid \vec{a} \right\mid \left\mid \vec{a} \right\mid = \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \left(C\right) + \left\mid \vec{a} \right\mid \left\mid \vec{c} \right\mid \cos \left(B\right)$

$\implies \left\mid \vec{a} \right\mid = \left\mid \vec{b} \right\mid \cos \left(C\right) + \left\mid \vec{c} \right\mid \cos \left(B\right)$