Cross Product — Question 10

Section 12-4 · Calculus pages 2379–2380

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Question 10

Vectors u→\vec u and v→\vec v have lengths 55 and 88, and ∥u→×v→∥=20\|\vec u\times\vec v\|=20. Find every possible angle between them in [0,π][0,\pi].

Question 10 – Solution

Keep the component order and signs organized when expanding the determinant. After computing the cross product, use a dot-product check or the relevant magnitude formula to interpret it.

Diagram shown in the original SVG worksheet.

Use ∥u→×v→∥=∥u→∥∥v→∥sin⁡θ\|\vec u\times\vec v\|=\|\vec u\|\|\vec v\|\sin\theta: 20=40sin⁡θ20=40\sin\theta.

Thus sin⁡θ=1/2\sin\theta=1/2.

On [0,π][0,\pi], the two solutions are θ=π/6 or 5π/6\boxed{\theta=\pi/6\text{ or }5\pi/6}. The cross-product magnitude cannot distinguish acute from obtuse orientation.

The result follows from the defining vector formulas used above, and each component, magnitude, or scalar condition has been checked against the information in the question.

The cross product produces a vector perpendicular to both inputs. Its direction follows the right-hand rule, while its magnitude records the area of the spanned parallelogram.