Cross Product — Question 1

Section 12-4 · Calculus pages 2361–2362

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Question 1

Compute a→×b→\vec a\times\vec b for a→=⟨1,2,−1⟩\vec a=\langle1,2,-1\rangle and b→=⟨3,0,4⟩\vec b=\langle3,0,4\rangle, then verify it is perpendicular to both vectors.

Question 1 – 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.

Expand the determinant: a→×b→=⟨2(4)−(−1)(0),(−1)(3)−1(4),1(0)−2(3)⟩=⟨8,−7,−6⟩\vec a\times\vec b=\langle2(4)-(-1)(0),\,(-1)(3)-1(4),\,1(0)-2(3)\rangle=\langle8,-7,-6\rangle.

Check (a→×b→)⋅a→=8−14+6=0(\vec a\times\vec b)\cdot\vec a=8-14+6=0.

Also (a→×b→)⋅b→=24−24=0(\vec a\times\vec b)\cdot\vec b=24-24=0, so a→×b→=⟨8,−7,−6⟩\boxed{\vec a\times\vec b=\langle8,-7,-6\rangle} is perpendicular to both.

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.