# Question #848d3

May 21, 2016

Given $\vec{A} = {A}_{x} \hat{i} + {A}_{y} \hat{j} + {A}_{z} \hat{k}$

So
$\vec{A} . \vec{A} = \left({A}_{x} \hat{i} + {A}_{y} \hat{j} + {A}_{z} \hat{k}\right) \cdot \left({A}_{x} \hat{i} + {A}_{y} \hat{j} + {A}_{z} \hat{k}\right)$

$\implies | \vec{A} | | . \vec{A} | \cos 0 = \left({A}_{x} \hat{i} \cdot {A}_{x} \hat{i} + {A}_{y} \hat{j} \cdot {A}_{x} \hat{i} + {A}_{z} \hat{k} \cdot {A}_{x} \hat{i} + {A}_{x} \hat{i} \cdot {A}_{y} \hat{j} + {A}_{y} \hat{j} \cdot {A}_{y} \hat{j} + {A}_{z} \hat{k} \cdot {A}_{y} \hat{j} + {A}_{x} \hat{i} \cdot {A}_{z} \hat{k} + {A}_{y} \hat{j} \cdot {A}_{z} \hat{k} + {A}_{z} \hat{k} \cdot {A}_{z} \hat{k}\right) \ldots . . \left(1\right)$

We know
$\hat{i} \cdot \hat{i} = | \hat{i} | | \hat{i} | \cos 0 = 1 \times 1 \times 1 = 1$
similarly
$\hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$

We know
$\hat{i} \cdot \hat{j} = | \hat{i} | | \hat{j} | \cos 90 = 1 \times 1 \times 0 = 0$
similarly
$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

So we have from relation (1)

$| \vec{A} {|}^{2} = {A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}$

$\implies | A | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}}$