Describing A Mathematical Result

[Bashyboy]

Homework Statement

Let u be an arbitrary fixed unit vector and show that an vector b satisfies [itex]b^2 = (\vec{u} \cdot \vec{b}) + (\vec{u} \times \vec{b})^2[/itex] 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. [itex]\vec{u} \cdot \vec{b} = |u||b|\cos \theta[/itex]. 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[itex]\vec{b_{\vec{u}}}[/itex], such that [itex]\vec{b_{\vec{u}}} = c \vec{u}[/itex], and [itex]\vec{b} = \vec{b_{\vec{u}}} + \vec{b_{||}}[/itex], where [itex]\vec{b_{||}}[/itex] 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. [itex]\vec{u} \cdot \vec{b} = |u||b|\cos \theta[/itex]. 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[itex]\vec{b_{\vec{u}}}[/itex], such that [itex]\vec{b_{\vec{u}}} = c \vec{u}[/itex], and [itex]\vec{b} = \vec{b_{\vec{u}}} + \vec{b_{||}}[/itex], where [itex]\vec{b_{||}}[/itex] 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 [itex](\vec{u} \cdot \vec{b})[/itex] rather than [itex](\vec{u} \cdot \vec{b})^2[/itex]?

 

[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]

[itex](\vec{u} \cdot \vec{b})^2=b^2\cos^2{\theta}[/itex]
[itex](\vec{u} \times \vec{b})\cdot(\vec{u} \times \vec{b})=b^2\sin^2{\theta}[/itex]