(a) [2 marks]

Using 3D distance formula:
\( VB = \sqrt{(6-3)^2 + (8-4)^2 + (0-9)^2} \)
\( = \sqrt{9 + 16 + 81} = \sqrt{106} \approx 10.3 \) units
A1 for correct method, A1 for correct answer

(b) [4 marks]

1. Find VC: \( \sqrt{(6-3)^2 + (0-4)^2 + (0-9)^2} = \sqrt{106} \) M1
2. Find BC: \( \sqrt{(6-6)^2 + (0-8)^2 + (0-0)^2} = 8 \) M1
3. Apply cosine rule:
\( \cos \theta = \frac{VB^2 + VC^2 – BC^2}{2 \times VB \times VC} = \frac{106 + 106 – 64}{212} = \frac{37}{53} \) M1
4. Final angle: \( \theta = \cos^{-1}(\frac{37}{53}) \approx 45.7^\circ \) A1