# Doubt From Vectors

I will be thankful for proof!

expand it by the property

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[\vec{a}-\vec{d} \ \ \vec{b} -\vec{d} \ \ \vec{c}-\vec{d}]=0 \Rightarrow \vec{a}-\vec{d} , \vec{b}-\vec{d} , \vec{c}-\vec{d} are coplanar so that \vec{a}-\vec{d} = x(\vec{b}-\vec{d} )+ y(\vec{c}-\vec{d} )

or (x+y-1)\vec{d} =-\vec{a}+x \vec{b}+y \vec{c} or d =-\dfrac{\vec{a}}{x+y-1}+\dfrac{x\vec{b}}{x+y-1}+\dfrac{y\vec{c}}{x+y-1}

i.e. \vec{d} = \lambda_1 \vec{a} + \lambda_2 \vec{b}+\lambda_3 \vec{c} where \lambda_1+\lambda_2+\lambda_3=1

Now [\vec{d } \ \ \vec{a} \ \ \vec{b}] = [\lambda_1 \vec{a} + \lambda_2 \vec{b}+\lambda_3 \vec{c} \ \ \vec{a} \ \ \vec{b}] which by the properties of the box product becomes [\lambda_3 \vec{c} \ \ \vec{a} \ \ \vec{b}]=\lambda_3 [\vec{c} \ \ \vec{a} \ \ \vec{b}]

Similarly expanding the others we get that RHS =

(\lambda_1+\lambda_2+\lambda_3) [\vec{a} \ \ \vec{b} \ \ \vec{c}]= [\vec{a} \ \ \vec{b} \ \ \vec{c}]

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There is a geometric way of looking at it. The given condition means that A,B,C,D are co-planar.

Imagine one case where D lies inside \triangle ABC

Now consider the tetrahedron with O as apex. Then Vol of tetrahedron OABC we know is \dfrac{[\vec{a} \ \ \vec{b} \ \ \vec{c}]}{6}

But we can divide OABC into the tetrahedrons ODAB, ODBC, ODCA.

Thus \dfrac{[\vec{a} \ \ \vec{b} \ \ \vec{c}]}{6}=\dfrac{[\vec{d} \ \ \vec{a} \ \ \vec{b}]}{6}+\dfrac{[\vec{d} \ \ \vec{b} \ \ \vec{c}]}{6}+\dfrac{[\vec{d} \ \ \vec{c} \ \ \vec{a}]}{6} which yields the required identity

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