To solve the given problems, let's follow the steps systematically.

(a) Showing the Vectors are Coplanar

Three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are coplanar if the scalar triple product of the vectors is zero. The scalar triple product is given by $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$.

Let's compute this step-by-step:

1. Find the cross product $\mathbf{b} \times \mathbf{c}$:

$\mathbf{b} = \langle 1, 2, 1 \rangle, \quad \mathbf{c} = \langle 0, 1, 3 \rangle$

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 0 & 1 & 3 \end{vmatrix}$$ $$= \mathbf{i}(2 \cdot 3 - 1 \cdot 1) - \mathbf{j}(1 \cdot 3 - 1 \cdot 0) + \mathbf{k}(1 \cdot 1 - 2 \cdot 0)$$ $$= \mathbf{i}(6 - 1) - \mathbf{j}(3 - 0) + \mathbf{k}(1 - 0)$$ $$= \mathbf{i}(5) - \mathbf{j}(3) + \mathbf{k}(1)$$ $$= \langle 5, -3, 1 \rangle$$ 2. **Calculate the dot product $$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$$:** $$\mathbf{a} = \langle 2, 3, -1 \rangle, \quad \mathbf{b} \times \mathbf{c} = \langle 5, -3, 1 \rangle$$ $$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 2 \cdot 5 + 3 \cdot (-3) + (-1) \cdot 1$$ $$= 10 - 9 - 1$$ $$= 0$$ Since the scalar triple product is zero, the vectors $$\mathbf{a}$$, $$\mathbf{b}$$, and $$\mathbf{c}$$ are coplanar. ### (b) Express $$\mathbf{a}$$ in Terms of $$\mathbf{b}$$ and $$\mathbf{c}$$ We need to express $$\mathbf{a}$$ as a linear combination of $$\mathbf{b}$$ and $$\mathbf{c}$$: $$\mathbf{a} = x\mathbf{b} + y\mathbf{c}$$ This means: $$\langle 2, 3, -1 \rangle = x \langle 1, 2, 1 \rangle + y \langle 0, 1, 3 \rangle$$ Let's break this down into component form: $$\begin{cases} 2 = x \cdot 1 + y \cdot 0 \\ 3 = x \cdot 2 + y \cdot 1 \\ -1 = x \cdot 1 + y \cdot 3 \end{cases}$$ From the first equation: $$2 = x \quad \Rightarrow \quad x = 2$$ Substituting $$x = 2$$ into the second and third equations: $$3 = 2 \cdot 2 + y \cdot 1 \quad \Rightarrow \quad 3 = 4 + y \quad \Rightarrow \quad y = 3 - 4 \quad \Rightarrow \quad y = -1$$ $$-1 = 2 \cdot 1 + (-1) \cdot 3 \quad \Rightarrow \quad -1 = 2 - 3 \quad \Rightarrow \quad -1 = -1 \quad \text{(which is true)}$$ Therefore, the vector $$\mathbf{a}$$ can be expressed as: $$\mathbf{a} = 2\mathbf{b} - \mathbf{c}$$ ### Summary (a) The vectors $$\mathbf{a}$$, $$\mathbf{b}$$, and $$\mathbf{c}$$ are coplanar since their scalar triple product is zero. (b) The vector $$\mathbf{a}$$ can be expressed in terms of $$\mathbf{b}$$ and $$\mathbf{c}$$ as: $$\mathbf{a} = 2\mathbf{b} - \mathbf{c}$$ Would you like any additional details or have any questions? Here are 5 related questions you might be interested in: 1. How to compute the dot product of two vectors? 2. What is the geometric interpretation of the cross product? 3. How to determine if three points are collinear using vectors? 4. What is the significance of the scalar triple product in 3D geometry? 5. Can a set of more than three vectors be coplanar? **Tip:** When working with vectors, always ensure your operations (dot product, cross product) are performed in the correct sequence to avoid errors.