# bb(ul hata), bb(ul hatb) and bb(ul hatc) are unit vectors, and the angle between bb(ul hatb) and bb(ul hatc) is pi/6. Show that  bb(ul hata) = +-2 (bb(ul hatb) xx bb(ul hatc))?

Apr 28, 2017

For Proof, refer to the Explanation.

#### Explanation:

Given that,

for unit vectors veca & vec b, veca.vecb=0, &, veca.vecc=0.

$\therefore \vec{a} \bot \vec{b} , \mathmr{and} , \vec{a} \bot \vec{c} , \text{ i.e., to say,}$

$\vec{a} \text{ is "bot "both } \vec{b} \mathmr{and} \vec{c} .$

Clearly, $\vec{a}$ is along $\vec{b} \times \vec{c} .$

Therefore, let us suppose that,

veca=k(vecbxxvecc) ; kne0............(ast)

$\Rightarrow | | \vec{a} | | = | | k \left(\vec{b} \times \vec{c}\right) | |$

$\Rightarrow | | \vec{a} | | = | k | | | \vec{b} | | | | \vec{c} | | \sin \angle \left(\vec{b} , \vec{c}\right)$

$\Rightarrow 1 = | k | \cdot 1 \cdot 1 \cdot \sin \left(\frac{\pi}{6}\right)$

$\Rightarrow | k | = 2 , \mathmr{and} , k = \pm 2.$

Hence, by $\left(\ast\right) , \vec{a} = \pm 2 \left(\vec{b} \times \vec{c}\right) .$

This completes the Proof.

Enjoy Maths.!

Apr 28, 2017

$\boldsymbol{\underline{\hat{a}}}$, $\boldsymbol{\underline{\hat{b}}}$ and $\boldsymbol{\underline{\hat{c}}}$ are unit vectors and so:

$| | \boldsymbol{\underline{\hat{a}}} | | = | | \boldsymbol{\underline{\hat{b}}} | | = | | \boldsymbol{\underline{\hat{c}}} | | = 1$

We know that:

$| | \boldsymbol{\underline{U}} \times \boldsymbol{\underline{V}} | | = | | \boldsymbol{\underline{U}} | | \cdot | | \boldsymbol{\underline{V}} | | \cdot \sin \theta$

And so

$| | \boldsymbol{\underline{b}} \times \boldsymbol{\underline{c}} | | = | | \boldsymbol{\underline{b}} | | \setminus | | \boldsymbol{\underline{c}} | | \setminus \sin \left(\frac{\pi}{6}\right)$
$\text{ } = 1 \cdot 1 \cdot \frac{1}{2}$
$\text{ } = \frac{1}{2}$

We also have:

$\boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{b}}} = 0 \implies \boldsymbol{\underline{\hat{a}}}$ perpendicular to $\boldsymbol{\underline{\hat{b}}}$
$\boldsymbol{\underline{\hat{a}}} \cdot \boldsymbol{\underline{\hat{c}}} = 0 \implies \boldsymbol{\underline{\hat{a}}}$ perpendicular to $\boldsymbol{\underline{\hat{c}}}$

We know that the cross product of any two vectors is perpendicular to both those vectors, and so it must be that:

$\boldsymbol{\underline{\hat{a}}} = l a m \mathrm{da} \left(\boldsymbol{\underline{\hat{b}}} \times \boldsymbol{\underline{\hat{c}}}\right)$

And it follows that also:

$| | \boldsymbol{\underline{\hat{a}}} | | = | | l a m \mathrm{da} \left(\boldsymbol{\underline{\hat{b}}} \times \boldsymbol{\underline{\hat{c}}}\right) | |$
$\therefore 1 = | l a m \mathrm{da} | \setminus | | \hat{b} \times \hat{c} | |$
$\therefore 1 = | l a m \mathrm{da} | \setminus \frac{1}{2}$
$\therefore | l a m \mathrm{da} | = 2 \implies l a m \mathrm{da} = \pm 2$

And therefore:

$\boldsymbol{\underline{\hat{a}}} = \pm 2 \left(\boldsymbol{\underline{\hat{b}}} \times \boldsymbol{\underline{\hat{c}}}\right) \setminus \setminus$ QED