Question

For three non zero vectors a b c prove that a-b b-c c-a =0

Answer 1

$\textbf{Concept:}$

$\text{Scalar triple product:}$

$[\vec{a}\;\;\vec{b}\;\;\vec{c}]=(\vec{a}{\times}\vec{b}).\vec{c}$

$\text{Now,}$

$[\vec{a}-\vec{b}\;\;\vec{b}-\vec{c}\;\;\vec{c}-\vec{a}]$

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

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

$\text{using}$

$\text{For any non zero vectro \vec{a} , \vec{a}\times\vec{a}=\vec{0} }$

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

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

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

$=[\vec{a}\;\;\vec{b}\;\;\vec{c}]-[\vec{a}\;\;\vec{b}\;\;\vec{a}]-[\vec{a}\;\;\vec{c}\;\;\vec{c}]+[\vec{a}\;\;\vec{c}\;\;\vec{a}]+[\vec{b}\;\;\vec{c}\;\;\vec{c}]-[\vec{b}\;\;\vec{c}\;\;\vec{a}]$

$\text{We know that}$

$\text{Scalar triple product vanishes if any two of them are equal}$

$=[\vec{a}\;\;\vec{b}\;\;\vec{c}]-[\vec{b}\;\;\vec{c}\;\;\vec{a}]$

$=[\vec{a}\;\;\vec{b}\;\;\vec{c}]-[\vec{a}\;\;\vec{b}\;\;\vec{c}]$

$=0$

$\therefore\bf[\vec{a}-\vec{b}\;\;\vec{b}-\vec{c}\;\;\vec{c}-\vec{a}]=0$

Answer 2

It has been proven that [ a-b, b-c, c-a ] = 0.

Given

To prove that [ a-b, b-c, c-a ] = 0

It consists of non zero vectors a, b, c

                   Scalar triple product,

                    [ a b c ] = ( a × b ) . c

Where,         a =  a-b

                    b =  b-c

                    c  =  c-a

                    [ a b c ] = ( a × b ) . c

                [ a-b, b-c, c-a ] =  { ( a-b ) × ( b-c ) } . ( c-a )

                                         = { ( ab - ac - $b^{2}$ + bc ) } . ( c-a )

                In scalar triple product, for any non zero, $b^{2}$ = b × b =0

                Squaring terms are zero, so neglect it.

                  [ a-b, b-c, c-a ] = { ( ab - ac + bc ) } . ( c-a )

                  = { ( ab ) . c - ( ac ) . c + ( bc ) . c - ( ab ) . a + ( ac ) . a

                                                 - ( bc ) . a }

                 = [ abc - acc + bcc - aab + aac - abc ]

                Squaring terms becomes zero,

                   [ a-b, b-c, c-a ] = [ abc - abc ]

                                           = 0.

Hence, proved that [ a-b, b-c, c-a ] = 0.

To learn more...

1. brainly.in/question/806994

2. brainly.in/question/14670974