- #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?

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- #1

- 14

- 0

u (dot) [ u x v]

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?

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- #3

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Hello Lurid. Welcome to PF !## 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?

You have two different kinds of

What is the result of taking the dot product (scalar product) of two vectors which are perpendicular to each other?

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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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