Relative to a fixed origin O, the point A has position vector $(2i - j + 5k)$, the point B has position vector $(5i + 2j + 10k)$, and the point D has position vector $(-i + j + 4k)$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 8

The vector equation of the line l can be expressed using point A and the direction vector $\overrightarrow{AB}$:

$\mathbf{r} = \mathbf{A} + t \overrightarrow{AB}$

Substituting for point A and $\overrightarrow{AB}$:

$\mathbf{r} = (2i - j + 5k) + t(3i + 3j + 5k)$

Thus, the vector equation is:

$\mathbf{r} = (2 + 3t)i + (-1 + 3t)j + (5 + 5t)k$

To find the angle BAD, we first calculate $\overrightarrow{AD}$:

$\overrightarrow{AD} = \overrightarrow{D} - \overrightarrow{A} = (-i + j + 4k) - (2i - j + 5k)$

This gives us:

$\overrightarrow{AD} = (-3i + 2j - k)$

Next, we use the cosine rule:

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

Calculating the dot product:

$\overrightarrow{AB} \cdot \overrightarrow{AD} = (3)(-3) + (3)(2) + (5)(-1) = -9 + 6 - 5 = -8$

Finding the magnitudes:

$|\overrightarrow{AB}| = \sqrt{3^2 + 3^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43}$

$|\overrightarrow{AD}| = \sqrt{(-3)^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$

Substituting these into the cosine rule:

$\cos \theta = \frac{-8}{\sqrt{43} \cdot \sqrt{14}}$

Calculating:

From this, we find:\n$\theta \approx 109^{\circ}$

To find the position vector of point C, we can use the relationship:

$\overrightarrow{C} = \overrightarrow{B} + \overrightarrow{AD}$

Substituting in our vectors:

$\overrightarrow{C} = (5i + 2j + 10k) + (-3i + 2j - k)$

Simplifying this gives:

$\overrightarrow{C} = (5 - 3)i + (2 + 2)j + (10 - 1)k = 2i + 4j + 9k$

The area of the parallelogram can be calculated using the formula:

$\text{Area} = |\overrightarrow{AB} \times \overrightarrow{AD}|$

Calculating the cross product:

$\overrightarrow{AB} = (3, 3, 5)$ $\overrightarrow{AD} = (-3, 2, -1)$

The determinant of the matrix formed gives:

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 5 \\ -3 & 2 & -1 \end{vmatrix}$

Calculating leads to:

$\text{Area} = 23.2$

To find the shortest distance from point D to line l, we use:

$d = \frac{|\overrightarrow{AD} \cdot \overrightarrow{n}|}{|\overrightarrow{n}|}$

Where $\overrightarrow{n}$ is a vector parallel to the line, calculated as:

$\overrightarrow{n} = \overrightarrow{AB}$

After substituting the values:

$d = \frac{sine(109^{\circ})}{\sqrt{14}}$

Calculating yields:

$d \approx 3.54$