# Dot and Cross Products

u (dot) [ u x v]

## 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?

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

SammyS
Staff Emeritus
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u (dot) [ u x v]

## 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?

Ray Vickson
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u (dot) [ u x v]

## 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 $\textbf{u} = u_x \textbf{i} + u_y\, \textbf{j} + u_z\textbf{k} \mbox{ and } \textbf{v} = v_x \textbf{i} + v_y \, \textbf{j} + v_z \textbf{k},$ then the x-component of $\textbf{u} \times \textbf{v}$ is $u_y v_z - u_z v_y.$ When we take the dot product with u, we take $u_x$ 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