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Hit this in my book. Can someone help me out?

Let $$ \begin{align} u &= i - j \\ v &= 2i - 3j + k. \end{align} $$

Find the area of the parallelogram defined by $u$ and $v$.

Am I right to take a 3rd point $v \times u$?

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You are on the right track. The magnitude of the cross-product gives the area. Do you know how to compute $v \times u$? – Sammy Black May 12 '13 at 8:05
It's best to think about $v \times u$ as a vector (well, a pseudovector) rather than a point. – Sammy Black May 12 '13 at 8:07
up vote 1 down vote accepted

The area of a parallelogram =$|\vec u \times\vec v |$ where $\vec u$ and $\vec v$ are vectors of conjugate sides of parallelogram. $\vec u=i-j$ and $\vec v=2i-3j+k$ $$\vec u \times \vec v= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\1 & -1 & 0 \\2 & -3 & 1 \end{vmatrix}$$ $$\vec u \times \vec v=\mathbf{i}(-1-0)-\mathbf{j}(1-0)+\mathbf{k}(-3+2)$$ $$\vec u \times \vec v=\mathbf{-i}-\mathbf{j}\mathbf{-k}$$ $$|\vec u \times\vec v |=\sqrt3$$ area of parallelogram is $\sqrt 3$

here $\mathbf{i}$,$\mathbf{j}$,$\mathbf{k}$ are unit vectors.

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