Hint
Use acute-angle conventions when requested. Normalise the absolute scalar product.
Worked solution
(a)
\(\displaystyle \frac{|a\cdot b|}{|a||b|}= \frac{4\sqrt{5330}}{2665}\)
Answer: \(0.109578760215799\).
Answer: \(\approx 0.109578760215799\)
(b)
\(\displaystyle \arccos (\frac{|a\cdot b|}{|a||b|})= 83.7089664806739\)
Answer: \(83.7089664806739\).
Answer: \(\approx 83.7089664806739\)
(c)
\(\displaystyle \frac{|Q\cdot (a\times b)|}{|a\times b|}= \frac{721\sqrt{5266}}{5266}\)
Answer: \(9.93561689575068\).
Answer: \(\approx 9.93561689575068\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Normalise the absolute scalar product.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: The line-plane angle is complementary to the line-normal angle.
(b) · 2 marks
- 1 mark: Use inverse cosine and report degrees.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: A dot product alone does not identify a shortest joining segment.
(c) · 2 marks
- 1 mark: Project the point displacement onto the common perpendicular.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: The line-plane angle is complementary to the line-normal angle.
Total: 6 marks.