Dot and Cross Products

  • Thread starter Lurid
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  • #1
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Homework Statement



u (dot) [ u x v]

Homework Equations





The Attempt at a Solution



The answer is 0, but I'm not sure why. Do you simplify [u x v], then use that vector to dot it with u? Or can you distribute the u to both u and v?
 

Answers and Replies

  • #2
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The vector of [tex]\left|u \times v\right| = \left|\begin{array}{cc}i & j & k\\u_1 & u_2 & u_3\\v_1 & v_2 &v_3\end{array}\right|[/tex]Then substitute in [tex] u [/tex] for [tex] i,j,k [/tex] because you're trying to prove that the vector [tex] \left|u \times v\right| [/tex] is orthogonal to both u and v. Then simplify and distribute.
 
  • #3
SammyS
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Homework Statement



u (dot) [ u x v]

Homework Equations



The Attempt at a Solution



The answer is 0, but I'm not sure why. Do you simplify [u x v], then use that vector to dot it with u? Or can you distribute the u to both u and v?
Hello Lurid. Welcome to PF !

You have two different kinds of multiplication involving vectors. Why would you be tempted to do any distributing ?


u × v is perpendicular to both u and v.

What is the result of taking the dot product (scalar product) of two vectors which are perpendicular to each other?
 
  • #4
Ray Vickson
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Homework Statement



u (dot) [ u x v]

Homework Equations





The Attempt at a Solution



The answer is 0, but I'm not sure why. Do you simplify [u x v], then use that vector to dot it with u? Or can you distribute the u to both u and v?

Sometimes the easiest way is to just write things down in detail. If [itex] \textbf{u} = u_x \textbf{i} + u_y\, \textbf{j} + u_z\textbf{k} \mbox{ and } \textbf{v} = v_x \textbf{i} + v_y \, \textbf{j} + v_z \textbf{k}, [/itex] then the x-component of [itex] \textbf{u} \times
\textbf{v}[/itex] is [itex] u_y v_z - u_z v_y. [/itex] When we take the dot product with u, we take [itex] u_x [/itex] times the above. We have two other such terms from the y- and z-components, so all we need to do is sum them up and simplify. It really is not difficult.

RGV
 

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