Find a vector perpendicular to another vector

  • Thread starter ch33zer
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  • #1
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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 [tex]\bullet[/tex] B = |A||B| cos(theta)
A [tex]\bullet[/tex] B = AxBx i + AyBy j + AzBz k
A [tex]\bullet[/tex] 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!
 
Last edited:

Answers and Replies

  • #2
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AXB is AB sin(theta)
 
  • #3
LCKurtz
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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.
 
  • #4
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Thanks. Got it.
 

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