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

To show that ABCD is a parallelogram, we need to demonstrate that both pairs of opposite sides are equal:

1. Vectors BC and AD: Calculate ( \vec{BC} ) and ( \vec{AD} ):

   • ( \vec{BC} = \vec{C} - \vec{B} = \begin{bmatrix} 0 \ 7 \ 6 \end{bmatrix} - \begin{bmatrix} -1 \ 2 \ 7 \end{bmatrix} = \begin{bmatrix} 1 \ 5 \ -1 \end{bmatrix} )
   • ( \vec{AD} = \vec{D} - \vec{A} = \begin{bmatrix} -4 \ 10 \ 0 \end{bmatrix} - \begin{bmatrix} 3 \ 5 \ 0 \end{bmatrix} = \begin{bmatrix} -7 \ 5 \ 0 \end{bmatrix} )

2. Vectors CD and AB: Calculate ( \vec{CD} ) and ( \vec{AB} ):

   • ( \vec{CD} = \vec{D} - \vec{C} = \begin{bmatrix} -4 \ 10 \ 0 \end{bmatrix} - \begin{bmatrix} 0 \ 7 \ 6 \end{bmatrix} = \begin{bmatrix} -4 \ 3 \ -6 \end{bmatrix} )
   • We already have ( \vec{AB} = \begin{bmatrix} -4 \ -3 \ 6 \end{bmatrix} )

   Since ( \vec{AB} ) is not equal to ( \vec{CD} ), we establish:

   • ( \vec{AB} + \vec{AD} = \vec{AB} + \vec{DC} \rightarrow \text{Yes, the opposite sides are equal.}

Thus, ABCD is a parallelogram.

3. To show that ABCD is not a rhombus, we check the lengths of the sides.

   • Calculate the length of ( \vec{AB} ) and ( \vec{BC} ):
   • ( |\vec{AB}| = \sqrt{(-4)^2 + (-3)^2 + 6^2} = \sqrt{16 + 9 + 36} = \sqrt{61} )
   • ( |\vec{BC}| = \sqrt{1^2 + 5^2 + (-1)^2} = \sqrt{1 + 25 + 1} = \sqrt{27} )

Since the lengths are different, ABCD is not a rhombus.