Automoderated users, edit
3,552
edits
Changes
Jump to navigation
Jump to search
Line 37:
Line 37: − + + + + + + +
Calc3.2.CrossProductProof (view source)
Revision as of 05:16, February 28, 2010
, 05:16, February 28, 2010→Third Identity
===Third Identity===
===Third Identity===
The final identity we will require takes a little bit more finesse. Suppose we have three vectors, <math>\vec{p},\vec{q},\vec{r} \ </math>. Suppose <math>\vec{p}\neq\vec{0} \ </math>. Let <math>P \ </math> be the plane through the origin which is perpendicular to <math>\vec{p} \ </math> (ie, the plane <math>p_1x+p_2y+p_3z=0</math>). If <math>\vec{q} \ </math> is any vector, let <math>\vec{q}_*</math> be the projection of <math>\vec{q}</math> into <math>P \ </math>.
The final identity we will require takes a little bit more finesse. Suppose we have three vectors, <math>\vec{p},\vec{q},\vec{r} \ </math>. Suppose <math>\vec{p}\neq\vec{0} \ </math>. Let <math>P \ </math> be the plane through the origin which is perpendicular to <math>\vec{p} \ </math> (ie, the plane <math>p_1x+p_2y+p_3z=0</math>). If <math>\vec{q} \ </math> is any vector, let <math>\vec{q}_*</math> be the projection of <math>\vec{q}</math> into <math>P \ </math>. Clearly, <math>\left\|{ \vec{q}_* }\right\|=\left\|{ \vec{q} \ }\right\|\sin(\theta) \ </math>, and so <math>\vec{p}\times\vec{q}=\vec{p}\times\vec{q}_* \ </math>. The reader may also assure themselves if they wish that <math>(\vec{q}+\vec{r})_* = \vec{q}_*+\vec{r}_* \ </math>.
Hence, to prove the third identity, it would suffice to prove instead <math>\vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ </math>. The advantage is that the vectors <math>\vec{q}_*, \vec{r}_*, \ </math> and <math>\vec{q}_*+\vec{r}_* \ </math> are all either <math>\vec{0}</math> or perpendicular to <math>\vec{p} \ </math>.
If either <math>\vec{q}_*, \vec{r}_*=\vec{0}</math>, then the identity we seek is certainly true, and if they are collinear, then one is a multiple of the other and the identity we seek follows from application of our first two identities.
So, let us assume that <math>\vec{q}_*, \vec{r}_*</math> both have non-zero magnitude and are not collinear. If we imagine <math>\vec{p}</math> sticking out of our computer monitors, pointing towards the reader, then the vectors in questions all look something like the picture on the right.
==Notes==
==Notes==
{{reflist}}
{{reflist}}