Question

The vectors a-b,b-c,c -a are
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Answer 1

Question :-

$\rm \: The \: vectors \: \vec{a} - \vec{b}, \: \vec{b} - \vec{c}, \: \vec{c} - \vec{a} \: are - - - -$

$\large\underline{\sf{Solution-}}$

Consider

$\rm \: [\vec{a} - \vec{b} \: \: \vec{b} - \vec{c} \: \: \vec{c} - \vec{a}]$

can be rewritten as

$\rm \: = (\vec{a} - \vec{b}).\bigg((\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) \bigg)$

$\rm \: = (\vec{a} - \vec{b}).\bigg(\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \bigg)$

We know,

$\boxed{ \rm{ \:\vec{a} \times \vec{a} = 0 \: }}$

So, using this, we get

$\rm \: = (\vec{a} - \vec{b}).\bigg(\vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a} \bigg)$

$\rm \: =\vec{a}.(\vec{b} \times \vec{c}) - \vec{a}.(\vec{b} \times \vec{a}) + \vec{a}.(\vec{c} \times \vec{a}) - \vec{b}.(\vec{b} \times \vec{c}) + \vec{b}.(\vec{b} \times \vec{a}) - \vec{b}.(\vec{c} \times \vec{a})$

$\rm \: =\vec{a}.(\vec{b} \times \vec{c}) - 0 + 0 - 0 + 0 - \vec{b}.(\vec{c} \times \vec{a})$

$\rm \: =\vec{a}.(\vec{b} \times \vec{c}) - \vec{a}.(\vec{b} \times \vec{c})$

$\rm \: = \: 0$

Thus,

$\rm\implies \:\rm \: [\vec{a} - \vec{b} \: \: \vec{b} - \vec{c} \: \: \vec{c} - \vec{a}] = 0$

$\rm\implies \: \vec{a} - \vec{b}, \: \vec{b} - \vec{c}, \: \vec{c} - \vec{a} \: are \: coplanar$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:[\vec{a} \: \: \vec{b} \: \: \vec{c}] \: = \: \vec{a}.(\vec{b} \times \vec{c}) \: \: }}$

$\boxed{ \rm{ \: \:\vec{a}.(\vec{b} \times \vec{c}) = \vec{b}.(\vec{c} \times \vec{a}) = \vec{c}.(\vec{a} \times \vec{b}) }}$

$\boxed{ \rm{ \: \:[\vec{a} \: \: \vec{b} \: \: \vec{c}] = 0 \: \rm\implies \:\vec{a},\vec{b},\vec{c} \: are \: coplanar \: }}$

$\boxed{ \rm{ \: \:[\vec{a} \: \: \vec{a} \: \: \vec{b}] = 0 \: \: }}$