LinAlg4.2.10

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Find the vector area of a parallelogram whose diagonals are determined by the vectors \bar{a}=3\bar{i}+\bar{j}-2\bar{k},\bar{b}=\bar{i}-3\bar{j}+4\bar{k}\,

\bar{a}\times\bar{b}=\begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}\,

\bar{i}(4-6)-\bar{j}(2+2)+\bar{k}(-9-1)=-2\bar{i}-14\bar{j}-10\bar{k}\,

Therefore,area of the parallelogram = \frac{1}{2}|a\times b| =| \frac{1}{2}\sqrt{(-2)^2+(-14)^2+(-10)^2} |=\frac{1}{2}\sqrt{300}=5\sqrt{3}\, units.


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