Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.
Rule: cross product calculated by a determinant
Let
$\text{u}=\u27e8{u}_{1},{u}_{2},{u}_{3}\u27e9$ and
$\text{v}=\u27e8{v}_{1},{v}_{2},{v}_{3}\u27e9$ be vectors. Then the cross product
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ is given by
Using determinant notation to find
$\text{p}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{q}$
Let
$\text{p}=\u27e8\mathrm{-1},2,5\u27e9$ and
$\text{q}=\u27e84,0,\mathrm{-3}\u27e9.$ Find
$\text{p}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{q}.$
We set up our determinant by putting the standard unit vectors across the first row, the components of
$\text{u}$ in the second row, and the components of
$\text{v}$ in the third row. Then, we have
Use determinant notation to find
$\text{a}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{b},$ where
$\text{a}=\u27e88,2,3\u27e9$ and
$\text{b}=\u27e8\mathrm{-1},0,4\u27e9.$
The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a
parallelepiped . The following examples illustrate these calculations.
Finding a unit vector orthogonal to two given vectors
Let
$\text{a}=\u27e85,2,\mathrm{-1}\u27e9$ and
$\text{b}=\u27e80,\mathrm{-1},4\u27e9.$ Find a unit vector orthogonal to both
$\text{a}$ and
$\text{b}.$
The cross product
$\text{a}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{b}$ is orthogonal to both vectors
$\text{a}$ and
$\text{b}.$ We can calculate it with a determinant:
Thus,
$\u27e8\frac{7}{\sqrt{474}},\frac{\mathrm{-20}}{\sqrt{474}},\frac{\mathrm{-5}}{\sqrt{474}}\u27e9$ is a unit vector orthogonal to
$\text{a}$ and
$\text{b}.$
To use the cross product for calculating areas, we state and prove the following theorem.
Area of a parallelogram
If we locate vectors
$\text{u}$ and
$\text{v}$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by
$\Vert \text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\Vert $ (
[link] ).
Proof
We show that the magnitude of the cross product is equal to the base times height of the parallelogram.
Let
$P=\left(1,0,0\right),Q=\left(0,1,0\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}R=\left(0,0,1\right)$ be the vertices of a triangle (
[link] ). Find its area.
We have
$\overrightarrow{PQ}=\u27e80-1,1-0,0-0\u27e9=\u27e8\mathrm{-1},1,0\u27e9$ and
$\overrightarrow{PR}=\u27e80-1,0-0,1-0\u27e9=\u27e8\mathrm{-1},0,1\u27e9.$ The area of the parallelogram with adjacent sides
$\overrightarrow{PQ}$ and
$\overrightarrow{PR}$ is given by
$\Vert \overrightarrow{PQ}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\overrightarrow{PR}\Vert \text{:}$
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