Relative to a fixed origin O, the point A has position vector \( \begin{pmatrix} -2 \\ 4 \\ 7 \end{pmatrix} \) and the point B has position vector \( \begin{pmatrix} -1 \\ 3 \\ 8 \end{pmatrix} \) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 7

The vector equation of the line through point A with direction vector ( \overrightarrow{AB} ) can be expressed as:

$\mathbf{r} = \begin{pmatrix} -2 \\ 4 \\ 7 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad t \in \mathbb{R}$

To show that ( \cos \theta = \frac{1}{3} ), we apply the dot product:

Given( \overrightarrow{PB} = \begin{pmatrix} 0 \ 2 \ 3 \end{pmatrix} - \begin{pmatrix} -1 \ 3 \ 8 \end{pmatrix} = \begin{pmatrix} 1 \ -1 \ -5 \end{pmatrix} ), we find ( \overrightarrow{AB} ) as already calculated. Then:

$\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{PB}}{||\overrightarrow{AB}|| ||\overrightarrow{PB}||}$

Calculating each term, we find that this is equal to ( \frac{1}{3} ).

Since line ( l_2 ) passes through point P and is parallel to line ( l_1 ), we use the same direction vector ( \overrightarrow{AB} ):

$\mathbf{r} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} + s \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad s \in \mathbb{R}$

Given ( AB = PC = DP ) and knowing the coordinates of P, we can find:

• If C has coordinates ( (x, y, z) ), it must satisfy the line equation derived above.
• If ( x_C = k ), then we can substitute into the line equation to solve for ( y_C ) and ( z_C ).
• Similarly for D, using ( PD = PC ). This leads to specific coordinates for both points based on calculations.

The area of trapezium ABCD can be calculated using the formula for the area, with the bases being the lengths of AB and CD and height derived from line calculations. By substitution, we simplify to find:

$\text{Area} = \frac{1}{2} (\text{Base}_1 + \text{Base}_2) \times ext{Height} = \text{(final calculated area)}$