How can the area of the parallelogram formed by two 3D vectors $$\mathbf{u} = [a, b, c], \space \mathbf{v} = [e, f, g]$$ be computed without using the cross product?
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2You may perhaps calculate the angle $\alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|\sin\alpha$, but it is basically the ugly way to calculate the cross product. – A.Γ. Jan 02 '19 at 20:22
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Rewrite $\mathbf v$ as the sum of two components, one parallel to $\mathbf u$ and one orthogonal to $\mathbf u.$ Let $\mathbf v_\perp$ be the orthogonal component. Then the area of the parallelogram is $$ \lVert u \rVert\, \lVert \mathbf v_\perp \rVert. $$
Projection of a vector and its answers explain how to get $\mathbf v_\parallel,$ the parallel component of $\mathbf v.$ And then $\mathbf v_\perp = \mathbf v - \mathbf v_\parallel.$ The effect is the same as the formula in the helpful comment under this answer.
David K
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How can I find the orthogonal component $\mathbf{v}_\perp$ or the length of the component directly? – K. Claesson Jan 02 '19 at 20:31
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2@K.Claesson Subtract from $\bf v$ the component of $\bf v$ along $\bf u$. $$\mathbf{v_\perp}=\mathbf v-\frac{\mathbf{v\cdot u}}{|\mathbf u|^2}\mathbf u$$ – Shubham Johri Jan 02 '19 at 20:41
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The square of the area is the determinant $$\begin{vmatrix} \bf{u}\cdot\bf u&\bf{u}\cdot\bf v\\ \bf{v}\cdot\bf u& \bf{v}\cdot\bf v \end{vmatrix}.$$
Angina Seng
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2@K.Claesson Recall that $A^2=|\mathbf u|^2|\mathbf v|^2\sin^2\alpha=|\mathbf u|^2|\mathbf v|^2(1-\cos^2\alpha)=|\mathbf u|^2|\mathbf v|^2-(\mathbf{u\cdot v})^2$ – Shubham Johri Jan 02 '19 at 20:49