If the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are coplanar, then find the following determinant

Question

If the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are coplanar, then find \begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \\ \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} & \vec{a}\cdot\vec{c} \\\vec{b}\cdot\vec{a} & \vec{b}\cdot\vec{b} & \vec{b}\cdot\vec{c}\\ \end{vmatrix}

I am not able to understand how to proceed to solve this determinant as it has dot products. Any idea on how to proceed? The answer is given as $0$.

Thanks.

Answer 1

If $\vec{a}, \vec{b}, \vec{c}$ are coplanar. This means there exists scalars $x,y,z$ (not all zero) such that $$x\vec{a}+y \vec{b}+z \vec{c}=\vec{0}.$$ Now $$(x\vec{a}+y \vec{b}+z \vec{c}) \cdot \vec{a}=\vec{0} \cdot \vec{a}=0$$ This gives $$x (\vec{a}\cdot \vec{a})+y (\vec{b}\cdot \vec{a})+z (\vec{c}\cdot \vec{a})=0$$ Likewise $$x (\vec{a}\cdot \vec{b})+y (\vec{b}\cdot \vec{b})+z (\vec{c}\cdot \vec{b})=0$$

Now you have the following homogeneous system of equations in $x,y,z$ such that it has non-zero solution. \begin{align*} x\vec{a}+y \vec{b}+z \vec{c} & =\vec{0}\\ x (\vec{a}\cdot \vec{a})+y (\vec{b}\cdot \vec{a})+z (\vec{c}\cdot \vec{a}) & =0\\ x (\vec{a}\cdot \vec{b})+y (\vec{b}\cdot \vec{b})+z (\vec{c}\cdot \vec{b})&=0. \end{align*}

Now think about determinants.