5. Cross Product
5. Geometric Interpretation of Cross Products
Like the dot product, there is also a geometric description of the cross product in terms of the magnitudes of the vectors \(\vec u\) and \(\vec v\) and the angle \(\theta\) between them. Unlike the dot product, which is a scalar, the cross product is a vector. So we need to give a geometrical description of both its magnitude and direction.
Let \(\vec u\) and \(\vec v\) be arbitrary vectors and let \(\theta\) be the angle between them. Then,
 The magnitude of the cross product may be expressed as \[ \vec u\times\vec v=\vec u\,\vec v\sin\theta \]
Recall the Pythagorean Identity for Dot and Cross Products : \[ (\vec u\cdot\vec v)^2+\vec u\times\vec v^2 =\vec u^2\vec v^2 \] Solve for \(\vec u\times\vec v^2\) and substitute the geometric formula for the dot product, \(\vec u\cdot\vec v=\vec u\,\vec v\cos\theta\): \[\begin{aligned} \vec u\times\vec v^2 &=\vec u^2\vec v^2(\vec u\cdot\vec v)^2 \\ &=\vec u^2\vec v^2\vec u^2\vec v^2\cos^2\theta \end{aligned}\] Next factor out the \(\vec u^2\vec v^2\) and use the Pythagorean identity from trig: \[\begin{aligned} \vec u\times\vec v^2 &=\vec u^2\vec v^2(1\cos^2\theta) \\ &=\vec u^2\vec v^2\sin^2\theta \end{aligned}\] Now take the square root of both sides\(:\) \[ \vec u\times\vec v=\vec u\,\vec v\,\sin\theta \] Note, although the lengths \(\vec u\), \(\vec v\) and \(\vec u\times\vec v\) are always nonnegative, \(\sin\theta\) may not be. So, we need an absolute value on \(\sin\theta\). However, the angle between two vectors is always between \(0\) (when they are in the same direction) and \(\pi\) (when they are in opposite directions), and for these values \(\sin\theta \ge 0\).
Consequently, we, in fact, don't need the absolute value. Therefore, \[ \vec u\times\vec v =\vec u\,\vec v\sin\theta \]

The direction of the cross product
\(\vec u\times\vec v\) is perpendicular to \(\vec u\) and to
\(\vec v\) and given by the right hand rule.
Be sure to read the two proofs!
Recall the Triple Product is zero if any two vectors are equal. Consequently: \[ \vec u\cdot\vec u\times\vec v=0 \qquad \text{and} \qquad \vec v\cdot\vec u\times\vec v=0 \]
These equations say \(\vec u\times\vec v\) is perpendicular to both \(\vec u\) and \(\vec v\).
To see that the cross product obeys the RHR, recall that \(\hat\imath\times\hat\jmath=\hat k\) satisfies the RHR.
Now let \(\hat\imath\) gradually rotate (and stretch) into \(\vec u\). Then \(\hat\imath\times\hat\jmath\) will gradually rotate (and stretch) into \(\vec u\times\hat\jmath\) continuously satisfing the RHR.
Finally let \(\hat\jmath\) gradually rotate (and stretch) into \(\vec v\). Then \(\vec u\times\hat\jmath\) will gradually rotate (and stretch) into \(\vec u\times\vec v\) continuously satisfing the RHR.
The
Pythagorean Identity for Dot and Cross Products
says:
\[
(\vec u\cdot\vec v)^2+\vec u\times\vec v^2
=\vec u^2\vec v^2
\]
Substitute the geometric formulas for the dot product and the magnitude
of the cross product:
\[
\vec u\cdot\vec v=\vec u\,\vec v\cos\theta
\qquad \text{and} \qquad
\vec u\times\vec v=\vec u\,\vec v\sin\theta
\]
into the Pythagorean identity:
\[
\vec u^2\vec v^2\cos^2\theta+\vec u^2\vec v^2\sin^2\theta
=\vec u^2\vec v^2
\]
So we see that the
Pythagorean Identity for Dot and Cross Products
is just \(\vec u^2\vec v^2\)
times the Pythagorean Identity for Sine and Cosine:
\[
\cos^2\theta+\sin^2\theta=1
\]
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