(i) To find the angle \(BOA\), use the scalar product formula:
\(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos \theta\).
Calculate the scalar product:
\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3)(5) + (4)(-2) + (-1)(-3) = 15 - 8 + 3 = 10\).
Calculate the magnitudes:
\(|\overrightarrow{OA}| = \sqrt{3^2 + 4^2 + (-1)^2} = \sqrt{26}\).
\(|\overrightarrow{OB}| = \sqrt{5^2 + (-2)^2 + (-3)^2} = \sqrt{38}\).
Substitute into the scalar product formula:
\(10 = \sqrt{26} \cdot \sqrt{38} \cdot \cos \theta\).
\(\cos \theta = \frac{10}{\sqrt{26} \cdot \sqrt{38}}\).
Calculate \(\theta\):
\(\theta = \cos^{-1}\left(\frac{10}{\sqrt{26} \cdot \sqrt{38}}\right) \approx 71.4\) or \(71.5\) degrees or \(1.25\) radians.
(ii) To find \(\overrightarrow{DC}\), first find \(\overrightarrow{OC}\):
\(\overrightarrow{OC} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}((3i + 4j - k) + (5i - 2j - 3k))\).
\(\overrightarrow{OC} = \frac{1}{2}(8i + 2j - 4k) = 4i + j - 2k\).
Given \(\overrightarrow{OD} = 2\overrightarrow{OB} = 2(5i - 2j - 3k) = 10i - 4j - 6k\).
Then \(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = (4i + j - 2k) - (10i - 4j - 6k)\).
\(\overrightarrow{DC} = -6i + 5j + 4k\).
