Cross Product — Question 9

PDF ↗

Question 9

Find a vector of magnitude 1010 perpendicular to the plane spanned by a→=⟨1,0,1⟩\vec a=\langle1,0,1\rangle and b→=⟨0,2,1⟩\vec b=\langle0,2,1\rangle, with positive zz-component.

Original worksheet page 1: question and worked solution for 5-4-009
Show solutionHide solution

Question 9 – 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.

See the diagram in the original worksheet below.

A normal direction is a→×b→=⟨−2,−1,2⟩\vec a\times\vec b=\langle-2,-1,2\rangle.

Its magnitude is 33, and its zz-component is positive.

Scale it to length 1010: n→=103⟨−2,−1,2⟩\boxed{\vec n=\frac{10}{3}\langle-2,-1,2\rangle}.

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.

Original worksheet page 2: question and worked solution for 5-4-009

Original worksheet layout. Use Enlarge or open the PDF for a closer view.