More practice questions

4. Q4. Find the area of the triangle with corners A (1, 0, 0), B (0, 1, 0) and C (0, 0, 1), in metres.

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   Two sides from A are AB = −$\hat i$ + $\hat j$ and AC = −$\hat i$ + $\hat k$. Their vector product has components (1)(1) − (0)(0) = 1, (0)(−1) − (−1)(1) = 1 and (−1)(0) − (1)(−1) = 1, so it is $\hat i$ + $\hat j$ + $\hat k$, of magnitude √3. The triangle’s area is half of that: √3/2, about 0.866 m^².

5. Q5. Show that |a × b|^² + (a · b)^² = a^²b^² for any two vectors.

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   If either vector is zero, both sides are zero. Otherwise, the left side is a^²b^² sin^²θ + a^²b^² cos^²θ = a^²b^²(sin^²θ + cos^²θ) = a^²b^².

6. Q6. For a = $\hat i$ + 2$\hat j$ and b = 2$\hat i$ + $\hat j$, find a × b and a · b, and use both to find the angle between the vectors.

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   a × b = (1 × 1 − 2 × 2)$\hat k$ = −3$\hat k$, of magnitude 3, and a · b = 2 + 2 = 4. Both vectors have magnitude √5, so ab = 5. Then sin θ = 3/5 and cos θ = 4/5, so θ is about 36.9°. (The sine alone would also allow 143.1°; the positive cosine rules it out.)