A quadrilateral has vertices A, B, C and D with position vectors given by: $$\vec{OA} = \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} -1 \\ 2 \\ 7 \end{pmatrix}, \quad \vec{OC} = \begin{pmatrix} 0 \\ 7 \\ 6 \end{pmatrix} \text{ and } \quad \vec{OD} = \begin{pmatrix} 4 \\ 10 \\ 0 \end{pmatrix}$ 14 (a) Write down the vector $\vec{AB}$ - AQA - A-Level Maths Mechanics - Question 14 - 2018 - Paper 2

First, we need to find the vectors for the other sides:

1. Calculate $\vec{BC}$:

$\vec{BC} = \vec{C} - \vec{B} = \begin{pmatrix} 0 \\ 7 \\ 6 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 7 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}$

2. Calculate $\vec{AD}$:

$\vec{AD} = \vec{D} - \vec{A} = \begin{pmatrix} 4 \\ 10 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix}$

We observe that:

$\vec{BC} = \vec{AD}$

This means that the opposite sides are equal; hence, ABCD is a parallelogram.

Now to check if it is a rhombus, we need to compare the lengths $|\vec{AB}|$ and $|\vec{BC}|$.

Calculate the length of $\vec{AB}$:

$|\vec{AB}| = \sqrt{(-4)^2 + (-3)^2 + 6^2} = \sqrt{16 + 9 + 36} = \sqrt{61}$

Now calculate the length of $\vec{BC}$:

$|\vec{BC}| = \sqrt{(1)^2 + (5)^2 + (-1)^2} = \sqrt{1 + 25 + 1} = \sqrt{27}$

Since $|\vec{AB}| \neq |\vec{BC}|$, ABCD is not a rhombus.

Thus, we have shown that ABCD is a parallelogram but not a rhombus.