Question
The quadrilateral ABCD has the following coordinates:
- \( A(1, 3, 5) \)
- \( B(4, 7, 5) \)
- \( C(5, 8, 7) \)
- \( D(2, 4, 7) \)
(a) Write down \( \overrightarrow{AD} \).
(b) Calculate:
- The size of \( \hat{B}AD \).
- The area of triangle \( BAD \).
(c) Show that ABCD is a parallelogram.
▶️ Answer/Explanation
Detailed Solution
(a) Finding \( \overrightarrow{AD} \)
\[ \overrightarrow{AD} = D – A = \begin{pmatrix} 2 – 1 \\ 4 – 3 \\ 7 – 5 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
(b)(i) Finding \( \hat{B}AD \) using the cosine rule
Using the scalar (dot) product formula:
\[ \cos \hat{B}AD = \frac{\overrightarrow{AB} \cdot \overrightarrow{AD}}{|\overrightarrow{AB}| |\overrightarrow{AD}|} \]
First, we find \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B – A = \begin{pmatrix} 4 – 1 \\ 7 – 3 \\ 5 – 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} \]
Now, calculating the dot product:
\[ \overrightarrow{AB} \cdot \overrightarrow{AD} = (3 \times 1) + (4 \times 1) + (0 \times 2) = 3 + 4 + 0 = 7 \]
Finding magnitudes:
\[ |\overrightarrow{AB}| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
\[ |\overrightarrow{AD}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
Applying the formula:
\[ \cos \hat{B}AD = \frac{7}{5 \times \sqrt{6}} \]
Solving for \( \hat{B}AD \):
\[ \hat{B}AD = \cos^{-1} \left( \frac{7}{5\sqrt{6}} \right) = 55.1^\circ (55.1417…) \]
(b)(ii) Finding the area of triangle \( BAD \)
Using the vector cross-product method:
\[ \text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AD} \right| \]
Computing the cross product:
\[ \overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 0 \\ 1 & 1 & 2 \end{vmatrix} \]
Expanding the determinant:
\[ = \mathbf{i} (4 \times 2 – 0 \times 1) – \mathbf{j} (3 \times 2 – 0 \times 1) + \mathbf{k} (3 \times 1 – 4 \times 1) \]
\[ = \mathbf{i} (8) – \mathbf{j} (6) + \mathbf{k} (-1) \]
\[ = (8, -6, -1) \]
Finding magnitude:
\[ \left| \overrightarrow{AB} \times \overrightarrow{AD} \right| = \sqrt{8^2 + (-6)^2 + (-1)^2} = \sqrt{64 + 36 + 1} = \sqrt{101} \]
\[ \text{Area} = \frac{1}{2} \times \sqrt{101} \approx 5.02 \]
(c) Proving ABCD is a parallelogram
We check if opposite sides are equal and parallel:
\[ \overrightarrow{AD} = \overrightarrow{BC} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
\[ \overrightarrow{AB} = \overrightarrow{DC} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} \]
Since both pairs of opposite sides are equal and parallel, quadrilateral ABCD is a parallelogram.
……………………………Markscheme……………………………….
(a)
\( \overrightarrow{AD} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)
(b) (i)
\( \hat{B}AD = 55.1^\circ \)
(ii)
\( \text{Area} = 5.02 \text{ units}^2 \)
(c)
\( \text{ABCD is a parallelogram as } \overrightarrow{AD} = \overrightarrow{BC} \text{ and } \overrightarrow{AB} = \overrightarrow{DC}. \)