Question

prove that a-b,b-c,c-a are coplaner​

Answer 1

$we know, three vector \vec{x},\vec{y} and \vec{z} are said to be coplanar when \vec{x}.(\vec{y}\times\vec{z}) </p><p></p><p>here, \vec{x}=\vec{a}-\vec{b}</p><p></p><p>\vec{y}=\vec{b}-\vec{c}</p><p></p><p>and \vec{z}=\vec{c}-\vec{a}</p><p></p><p>now, (\vec{a}-\vec{b}).\{(\vec{b}-\vec{c})\times(\vec{c}-\vec{a})\}</p><p></p><p>= (\vec{a}-\vec{b}).\{\vec{b}\times\vec{c}-\vec{b}\times\vec{a}-\vec{c}\times\vec{c}+\vec{c}\times\vec{a}\}</p><p></p><p>= (\vec{a}-\vec{b}).\{\vec{b}\times\vec{c}-\vec{b}\times\vec{a}+\vec{c}\times\vec{a}\}</p><p></p><p>= [\vec{a}\vec{b}\vec{c}]-0-0-0-0-[\vec{a}\vec{b}\vec{c}]</p><p></p><p>here, [\vec{a}\vec{b}\vec{c}] means, \vec{a}.(\vec{b}\times\vec{c})</p><p></p><p>= 0 </p><p></p><p>hence, \vec{a}-\vec{b},\vec{b}-\vec{c} and \vec{c}-\vec{a}$

Answer 2

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