Vector component perpendicular

[Mdhiggenz]

Homework Statement

Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then
projw(v) = < -2/3, 2/3, 2/3 > ,
and the component of v perpendicular to w is?

Homework Equations

The Attempt at a Solution

I understood how to get the projection, and tried using the component formula Compaonb= a*b/magnitude( A )

But did not get the correct answer which is w2 = < -7/3, -8/3, 1/3 >

Where did I go wrong?

 

[Muphrid]

Is that projection given? Because that doesn't jive with what I get.

If you find the part of v that is parallel to the direction of w, then the only other part of v that remains must be perpendicular to w, no?

 

[Mdhiggenz]

That is the answer to the projection yes, I don't understand what you are trying to explain though?

 

[Muphrid]

Okay, right, that projection is good. That is indeed the projection of $v$ onto $w$. You could call it $v_\parallel$.

The point I'm making is that, if $a + b = c$, then $b = c - a$, so if $v_\parallel + v_\perp = v$, then $v_\perp = ?$

 

[Mdhiggenz]

Perfect I got the correct answer. So what your saying is by getting the projection of v onto w. It gives me a vector V which is parallel to W, and to get the component of that vector we subtract the original vector v by the parallel component " if we want the perpendicular component".

Where I am confused is that there is a formula in my book for getting the component, and its in the form Comp v onto w = v*w/norm of v

Why would using that be incorrect?

Thank you for your help btw

 

[Chestermiller]

The component of v parallel to w is obtained by first calculating the unit vector in the direction of w (dividing w by its own magnitude), and then dotting v with that unit vector. The component perpendicular to w is obtained by subtracting the component parallel to w (times the unit vector in the direction of w) from v.