Given that vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, we need to show that the determinant of the following matrix is equal to zero: $\begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \end{vmatrix} = 0$

1. Problem Description

Given that vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, we need to show that the determinant of the following matrix is equal to zero:

$\begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \end{vmatrix} = 0$

2. Solution Steps

Since the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, one of them can be written as a linear combination of the other two. Let's assume $\vec{c} = x\vec{a} + y\vec{b}$, where $x$ and $y$ are scalars.

We can substitute this expression for $\vec{c}$ into the determinant:

$\begin{vmatrix} \vec{a} & \vec{b} & x\vec{a} + y\vec{b} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot (x\vec{a} + y\vec{b}) \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot (x\vec{a} + y\vec{b}) \end{vmatrix}$

Using the properties of the dot product, we have:

$\vec{a} \cdot (x\vec{a} + y\vec{b}) = x(\vec{a} \cdot \vec{a}) + y(\vec{a} \cdot \vec{b})$

$\vec{b} \cdot (x\vec{a} + y\vec{b}) = x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b})$

Substitute these back into the determinant:

$\begin{vmatrix} \vec{a} & \vec{b} & x\vec{a} + y\vec{b} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & x(\vec{a} \cdot \vec{a}) + y(\vec{a} \cdot \vec{b}) \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) \end{vmatrix}$

Now, we can split the third column into two columns using the property of determinants:

$\begin{vmatrix} \vec{a} & \vec{b} & x\vec{a} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & x(\vec{a} \cdot \vec{a}) \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & x(\vec{b} \cdot \vec{a}) \end{vmatrix} + \begin{vmatrix} \vec{a} & \vec{b} & y\vec{b} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & y(\vec{a} \cdot \vec{b}) \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & y(\vec{b} \cdot \vec{b}) \end{vmatrix}$

We can factor out the scalars $x$ and $y$ from the third columns:

$x\begin{vmatrix} \vec{a} & \vec{b} & \vec{a} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{a} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{a} \end{vmatrix} + y\begin{vmatrix} \vec{a} & \vec{b} & \vec{b} \\ \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix}$

Since the determinant of a matrix with two identical columns is zero, both determinants are zero.

$x(0) + y(0) = 0$

3. Final Answer

The determinant is equal to zero.