Hint
Position vectors use the same origin. Subtract the first position components.
Worked solution
(a)
Subtract b−a, then divide the nonzero displacement by its positive magnitude.
Answer: \(\left[\begin{matrix}2\\5\end{matrix}\right]\); \(\sqrt{29}\); \(\displaystyle \left[\begin{matrix}\frac{2 \sqrt{29}}{29}\\\frac{5 \sqrt{29}}{29}\end{matrix}\right]\)
(b)
Subtract the second position components.
\(5\)
Answer: \(5\)
(c)
Average the first components.
\(10\)
Answer: \(10\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 3 marks
- 1 mark: Subtract the endpoint position vectors in the required order.
- 1 mark: Find the positive magnitude and divide the displacement by it to obtain a unit vector.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: AB is b−a, not a−b.
(b) · 2 marks
- 1 mark: Subtract the second position components.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A direction vector is not generally a position vector.
(c) · 2 marks
- 1 mark: Average the first components.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: AB is b−a, not a−b.
Total: 7 marks.