Vectors — Area formula then equal-angle circles

Question

(i) Given that is perpendicular to with equal magnitude, show that the area of with and is .

(ii) Point lies on the circle centre radius with at angle to the horizontal. Point lies on the circle centre radius with also at angle to the horizontal. Using part (i), find that maximise the area of .

This practice question was inspired by Question 14(c) in the 2023 NSW HSC Mathematics Extension 1 examination, © NESA. It is an original TestMum variation that practises the same mathematical idea, not an official NESA question.

Step by step

Set up the solution: write the vector relationship component by component, then use a dot product, magnitude or scalar multiple as required.(i) Area . With perpendicular to and , .
Simplify the previous line carefully, keeping exact values where possible.Hence area .
Simplify the previous line carefully, keeping exact values where possible.(ii) , .
Finish the calculation, then check that the result meets the question’s conditions.Area .
State the final answer clearly in the form the question requested.On , is maximised at .

Common mistake: changing so many features that the new example no longer tests the same mathematical idea.

Verification: Full (i)–(ii): prove determinant area, then maximise with equal polar angles.

Any mark labels are a TestMum study aid, not an official NESA marking allocation.

← Back to 2023 HSC Extension 1 paper Question 14(c)