# If both If bb(ulhata) and bb(ulhat b) are unit vectors and  || bb(ul hata)-bb(ul hatb) || = sqrt(3)  show that  || bb(ul hata) + bb(ul hatb) || = 1?

Apr 28, 2017

To prove that $| \hat{a} + \hat{b} | = 1$.
Look for the explanation below.

#### Explanation:

Let $\hat{a} = c \hat{i} + \mathrm{dh} a t j + e \hat{k}$ and $\hat{b} = f \hat{i} + g \hat{j} + h \hat{k}$
where $\sqrt{{c}^{2} + {d}^{2} + {e}^{2}} = 1 = \sqrt{{f}^{2} + {g}^{2} + {h}^{2}}$

$\implies \textcolor{g r e e n}{{c}^{2} + {d}^{2} + {e}^{2} = 1 = {f}^{2} + {g}^{2} + {h}^{2}}$ ------------- 1.

$\implies \hat{a} - \hat{b} = \left(c - f\right) \hat{i} + \left(d - g\right) \hat{j} + \left(e - h\right) \hat{k}$

Since $| \hat{a} - \hat{b} | = \sqrt{3}$

$\implies \sqrt{{\left(c - f\right)}^{2} + {\left(d - g\right)}^{2} + {\left(e - h\right)}^{2}} = \sqrt{3}$

$\implies \sqrt{{c}^{2} + {f}^{2} - 2 c f + {d}^{2} + {g}^{2} - 2 \mathrm{dg} + {e}^{2} + {h}^{2} - 2 e h} = \sqrt{3}$

$\implies \sqrt{{c}^{2} + {d}^{2} + {e}^{2} + {f}^{2} + {g}^{2} + {h}^{2} - 2 \left(c f + \mathrm{dg} + e h\right)} = \sqrt{3}$

Using values of ${c}^{2} + {d}^{2} + {e}^{2}$ & ${f}^{2} + {g}^{2} + {h}^{2}$ from 1.

$\implies \sqrt{1 + 1 - 2 \left(c f + \mathrm{dg} + e h\right)} = \sqrt{3}$

squaring both sides

$\implies 2 - 2 \left(c f + \mathrm{dg} + e h\right) = 3$

$\implies \textcolor{g r e e n}{2 \left(c f + \mathrm{dg} + e h\right) = - 1}$ ---------- 2.

Now, the sum of $\hat{a} \mathmr{and} \hat{b}$

$\hat{a} + \hat{b} = \left(c + f\right) \hat{i} + \left(d + g\right) \hat{j} + \left(e + h\right) \hat{k}$

Let us find magnitude of $\hat{a} + \hat{b}$

$| \hat{a} + \hat{b} | = \sqrt{{\left(c + f\right)}^{2} + {\left(d + g\right)}^{2} + {\left(e + h\right)}^{2}}$

$= \sqrt{{c}^{2} + {f}^{2} + 2 c f + {d}^{2} + {g}^{2} + 2 \mathrm{dg} + {e}^{2} + {h}^{2} + 2 e h}$

$= \sqrt{{c}^{2} + {d}^{2} + {e}^{2} + {f}^{2} + {g}^{2} + {h}^{2} + 2 \left(c f + \mathrm{dg} + e h\right)}$

Using values of ${c}^{2} + {d}^{2} + {e}^{2}$ & ${f}^{2} + {g}^{2} + {h}^{2}$ from 1. and value of $2 \left(c f + \mathrm{dg} + e h\right)$ form 2.

$| \hat{a} + \hat{b} | = \sqrt{1 + 1 - 1} = \sqrt{1} = 1$

$\therefore \textcolor{red}{| \hat{a} + \hat{b} | = 1}$

Since the magnitude of $\hat{a} + \hat{b}$ is $1$, it is a unit vector.
$\therefore$ sum of $\hat{a}$ & $\hat{b}$ is a unit vector.

Apr 28, 2017

By definition then $| | \boldsymbol{\underline{A}} | {|}^{2} = \boldsymbol{\underline{A} \cdot \underline{A}}$

So, if both If $\boldsymbol{\underline{\hat{a}}}$ and $\boldsymbol{\underline{\hat{b}}}$ are unit vectors then:

$| | \boldsymbol{\underline{\hat{a}}} | | = 1 \iff \boldsymbol{\underline{a} \cdot \underline{a}} = 1$

$| | \boldsymbol{\underline{\hat{b}}} | | = 1 \iff \boldsymbol{\underline{b} \cdot \underline{b}} = 1$

And also we have:

$\boldsymbol{\underline{a} \cdot \underline{b}} = \boldsymbol{\underline{b} \cdot \underline{a}}$

We are given that $| | \boldsymbol{\underline{\hat{a}}} - \boldsymbol{\underline{\hat{b}}} | | = \sqrt{3}$, and so:

$| | \boldsymbol{\underline{\hat{a}}} - \boldsymbol{\underline{\hat{b}}} | {|}^{2} = 3$

But using the above definition we also have that:

 || bb(ul hata)-bb(ul hatb) ||^2 = (bb(ul hata)-bb(ul hatb) * (bb(ul hata)-bb(ul hatb))

$\therefore \left(\boldsymbol{\underline{\hat{a}}} - \boldsymbol{\underline{\hat{b}}}\right) \cdot \left(\boldsymbol{\underline{\hat{a}}} - \boldsymbol{\underline{\hat{b}}}\right) = 3$
$\therefore \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{a}}} - \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} - \boldsymbol{\underline{\hat{b}}} \cdot \boldsymbol{\underline{\hat{a}}} + \boldsymbol{\underline{\hat{b}}} \cdot \boldsymbol{\underline{\hat{b}}} = 3$
$\therefore \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{a}}} - 2 \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} + \boldsymbol{\underline{\hat{b}}} \cdot \boldsymbol{\underline{\hat{b}}} = 3$
$\therefore 1 - 2 \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} + 1 = 3$
$\therefore 2 \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} = - 1$

Similarly:

$| | \boldsymbol{\underline{\hat{a}}} + \boldsymbol{\underline{\hat{b}}} | {|}^{2} = \left(\boldsymbol{\underline{\hat{a}}} + \boldsymbol{\underline{\hat{b}}}\right) \cdot \left(\boldsymbol{\underline{\hat{a}}} + \boldsymbol{\underline{\hat{b}}}\right)$
$\text{ } = \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{a}}} + 2 \boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} + \boldsymbol{\underline{\hat{b}}} \cdot \boldsymbol{\underline{\hat{b}}}$
$\text{ } = 1 - 1 + 1$
$\text{ } = 1$

And so:

$| | \boldsymbol{\underline{\hat{a}}} + \boldsymbol{\underline{\hat{b}}} | | = 1 \setminus \setminus \setminus$ QED

Apr 28, 2017

Refer to the Explanation.

#### Explanation:

Knowing that,

$| | \hat{a} + \hat{b} | {|}^{2} + | | \hat{a} - \hat{b} | {|}^{2} = 2 \left[| | \hat{a} | {\setminus}^{2} + | | \hat{b} | {|}^{2}\right] ,$ we have, by what is

given, $| | \hat{a} + \hat{b} | {|}^{2} + {\left(\sqrt{3}\right)}^{2} = 2 \left[{1}^{2} + {1}^{2}\right] ,$

$\Rightarrow | | \hat{a} + \hat{b} | {|}^{2} = 4 - 3 = 1 , \mathmr{and} ,$

$| | \hat{a} + \hat{b} | | = 1 ,$ as desired.

Enjoy Maths.!