Describing A Mathematical Result

[Bashyboy]

Homework Statement

Let u be an arbitrary fixed unit vector and show that an vector b satisfies $b^2 = (\vec{u} \cdot \vec{b}) + (\vec{u} \times \vec{b})^2$ Explain this result in words, with the help of a picture.

Homework Equations

The Attempt at a Solution

I understand that the equations says that the square of the magnitude of some vector b is equal to the square of the dot product of b and some arbitrary unit vector u, plus the square of the cross product between the two vectors alluded to already.

I want to examine the dot product first. $\vec{u} \cdot \vec{b} = |u||b|\cos \theta$. Is it correct to state that the cross product represents the amount of vector b that goes (points) in the direction of vector u. So, the right side of the equation can be thought of the magnitude of some vector$\vec{b_{\vec{u}}}$, such that $\vec{b_{\vec{u}}} = c \vec{u}$, and $\vec{b} = \vec{b_{\vec{u}}} + \vec{b_{||}}$, where $\vec{b_{||}}$ is orthogonal to the vector u.

Are these correct statements?

 

[yands]

Can you write the question as it is supposed to be written.

 

[Bashyboy]

The description given in section 1 is the exact problem.

 

[Bashyboy]

I want to examine the dot product first. $\vec{u} \cdot \vec{b} = |u||b|\cos \theta$. Is it correct to state that the cross product represents the amount of vector b that goes (points) in the direction of vector u. So, the right side of the equation can be thought of the magnitude of some vector$\vec{b_{\vec{u}}}$, such that $\vec{b_{\vec{u}}} = c \vec{u}$, and $\vec{b} = \vec{b_{\vec{u}}} + \vec{b_{||}}$, where $\vec{b_{||}}$ is orthogonal to the vector u.

Are these correct statements?

I simply want to know if these statements are valid.

 

[Chestermiller]

Are you sure that the first term on the rhs is $(\vec{u} \cdot \vec{b})$ rather than $(\vec{u} \cdot \vec{b})^2$?

 

[Bashyboy]

Chester, you are correct. It should be squared.

 

[Bashyboy]

So, am I to assume the statements I quoted in post #4 are correct, as no one has opposed them?

 

[Chestermiller]

$(\vec{u} \cdot \vec{b})^2=b^2\cos^2{\theta}$
$(\vec{u} \times \vec{b})\cdot(\vec{u} \times \vec{b})=b^2\sin^2{\theta}$