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Worked solution

Find and verify a common perpendicular vector

Find a vector perpendicular to a=(211)\mathbf a=\begin{pmatrix}2\\1\\1\end{pmatrix} and b=(214)\mathbf b=\begin{pmatrix}2\\1\\4\end{pmatrix}.

CJC · Vectors Basic · Example 17

Answer

One such vector is (3−60)\begin{pmatrix}3\\-6\\0\end{pmatrix}; every nonzero scalar multiple also works.

Compare your reasoning

Take the cross product in the given order

(211)×(214)=((1×4)−(1×1)−[(2×4)−(1×2)](2×1)−(1×2))=(3−60)\begin{pmatrix}2\\1\\1\end{pmatrix}\times\begin{pmatrix}2\\1\\4\end{pmatrix}=\begin{pmatrix}(1\times4)-(1\times1)\\-[(2\times4)-(1\times2)]\\(2\times1)-(1\times2)\end{pmatrix}=\begin{pmatrix}3\\-6\\0\end{pmatrix}

The middle component includes the minus sign.

Verify both scalar products

(3−60)⋅(211)=6−6=0,(3−60)⋅(214)=6−6=0\begin{pmatrix}3\\-6\\0\end{pmatrix}\cdot\begin{pmatrix}2\\1\\1\end{pmatrix}=6-6=0,\qquad\begin{pmatrix}3\\-6\\0\end{pmatrix}\cdot\begin{pmatrix}2\\1\\4\end{pmatrix}=6-6=0

The printed remark recommends this independent check. Also check the result is nonzero: the zero vector would pass both dot tests without giving a direction.

How did this question go?

Solution notes

The Teacher fills the cross-product calculation; both copies print the two-dot-product checking remark.