Porisma

Worked solution

Calculate two cross products in the stated order

Find the vector product of (i) (117)\begin{pmatrix}1\\1\\7\end{pmatrix} and (5−42)\begin{pmatrix}5\\-4\\2\end{pmatrix}; (ii) 2i+4j+4k2\mathbf i+4\mathbf j+4\mathbf k and i+j−k\mathbf i+\mathbf j-\mathbf k.

CJC · Vectors Basic · Self-Practice 6

(i) Column vectors

Answer

(3033−9)\begin{pmatrix}30\\33\\-9\end{pmatrix}.

Keep the cyclic order

(117)×(5−42)=(1(2)−7(−4)7(5)−1(2)1(−4)−1(5))=(3033−9)\begin{pmatrix}1\\1\\7\end{pmatrix}\times\begin{pmatrix}5\\-4\\2\end{pmatrix}=\begin{pmatrix}1(2)-7(-4)\\7(5)-1(2)\\1(-4)-1(5)\end{pmatrix}=\begin{pmatrix}30\\33\\-9\end{pmatrix}

Working is authored; the final column is printed.

How did this part go?

(ii) Basis vectors

Answer

(−86−2)\begin{pmatrix}-8\\6\\-2\end{pmatrix}.

Read the three coefficients then cross

(244)×(11−1)=(4(−1)−4(1)4(1)−2(−1)2(1)−4(1))=(−86−2)\begin{pmatrix}2\\4\\4\end{pmatrix}\times\begin{pmatrix}1\\1\\-1\end{pmatrix}=\begin{pmatrix}4(-1)-4(1)\\4(1)-2(-1)\\2(1)-4(1)\end{pmatrix}=\begin{pmatrix}-8\\6\\-2\end{pmatrix}

Changing the vector order would negate every component.

How did this part go?

Solution notes

Both printed final columns are retained; full working is authored.