(i) To show that angle OAB is a right angle, we need to show that vectors OA and AB are perpendicular. The position vector of A is 3i + 2j - k and the position vector of B is 7i - 3j + k.
The vector AB is given by:
\(\mathbf{AB} = \mathbf{B} - \mathbf{A} = (7\mathbf{i} - 3\mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}\)
The dot product \(\mathbf{AO} \cdot \mathbf{AB}\) is:
\((3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \cdot (4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}) = 3 \times 4 + 2 \times (-5) + (-1) \times 2 = 12 - 10 - 2 = 0\)
Since the dot product is zero, angle OAB is a right angle.
(ii) To find the area of triangle OAB, we use the formula for the area of a triangle given by vectors:
\(\text{Area} = \frac{1}{2} \times |\mathbf{OA}| \times |\mathbf{AB}|\)
First, calculate the magnitudes:
\(|\mathbf{OA}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\)
\(|\mathbf{AB}| = \sqrt{4^2 + (-5)^2 + 2^2} = \sqrt{16 + 25 + 4} = \sqrt{45}\)
Thus, the area is:
\(\text{Area} = \frac{1}{2} \times \sqrt{14} \times \sqrt{45} = \frac{1}{2} \times \sqrt{630} = 12.5\)
