- #1
Cosmophile
- 111
- 2
Let ##\vec {A}## be an arbitrary vector and let ##\hat {n}## be a unit vector in some fixed direction. Show that [tex] \vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}. [/tex]
Here is my attempt, which I do believe is fine; I just want to know if there is a better way to approach the problem:
The first term, ##(\vec{A} \cdot \hat {n})\hat{n}## is a new vector ##\vec {C}## with magnitude ##A \cos \theta## in the ##\hat {n}## direction. For the second term, I simply did the operations from left-to-right, first getting ##\vec {C} = ( \hat {n} \times \vec {A})##, which if looking down on the plane made by ##\vec {A}## and ##\hat {n}##, would point upwards out of the page (I drew ##\vec {A} to be above \hat {n}##). Then, I took ##\vec {C} \times \hat {n}## and called this new vector ##\vec {D}## which has a magnitude ##A \sin \theta## and points in a direction ##\hat d##, which is perpendicular to ##\hat {n}##. So, I end up with [tex] \vec {A} = A(\cos \theta \hat {n} + \sin \theta \hat{d}). [/tex]
So, I've shown that the statement ##\vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}## just breaks ##\vec {A}## up into its horizontal and vertical components. At least, that's what I make of my work. What do you guys think?
Thanks in advance!
Here is my attempt, which I do believe is fine; I just want to know if there is a better way to approach the problem:
The first term, ##(\vec{A} \cdot \hat {n})\hat{n}## is a new vector ##\vec {C}## with magnitude ##A \cos \theta## in the ##\hat {n}## direction. For the second term, I simply did the operations from left-to-right, first getting ##\vec {C} = ( \hat {n} \times \vec {A})##, which if looking down on the plane made by ##\vec {A}## and ##\hat {n}##, would point upwards out of the page (I drew ##\vec {A} to be above \hat {n}##). Then, I took ##\vec {C} \times \hat {n}## and called this new vector ##\vec {D}## which has a magnitude ##A \sin \theta## and points in a direction ##\hat d##, which is perpendicular to ##\hat {n}##. So, I end up with [tex] \vec {A} = A(\cos \theta \hat {n} + \sin \theta \hat{d}). [/tex]
So, I've shown that the statement ##\vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}## just breaks ##\vec {A}## up into its horizontal and vertical components. At least, that's what I make of my work. What do you guys think?
Thanks in advance!