Find a vector perpendicular to another vector

[ch33zer]

Homework Statement

Consider the two vectors:
L = 4 i + 3 j + k
and
s = 6 i + 6 j + 8 k
Find the value of the scalar α such that the vector
L - αs
is perpendicular to L.

Homework Equations

Dot Product:
A \bullet B = |A||B| cos(theta)
A \bullet B = A_xB_x i + A_yB_y j + A_zB_z k
A \bullet A = (Ax^2 + Ay ^2 + Az^2)^.5 (Wouldn't let me do sub and sup in sqrt)

Cross Product
A x B = |A||B| sin(theta)

The Attempt at a Solution

I thought that I could do AxB and set that equal to |A||B| cos(theta) but when I did that everything just canceled out. I am really confused about multiplying a vector by a scalar and how that changes orientation etc.

Any help appreciated!

 

[cupid.callin]

AXB is AB sin(theta)

 

[LCKurtz]

First, AxB is not ABsin(θ). That is a statement about magnitudes:

|A x B| = |A||B| sin(θ)

although it is pretty much irrelevant to the question.

Calculate V = L - as and use the property that V is perpendicular to L if V · L = 0.

 

[ch33zer]

Thanks. Got it.