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.

 

[rwisz]

Hello! I can provide some guidance on how to find a vector perpendicular to another vector.

First, let's define what it means for two vectors to be perpendicular. Two vectors are perpendicular if their dot product is equal to zero. This means that they are at a 90 degree angle to each other.

In this problem, we are given two vectors, L and S, and we want to find a scalar, α, such that L - αS is perpendicular to L. To do this, we can use the dot product formula:

L \bullet (L - αS) = |L||L - αS| cos(theta)

Since we want the result to be perpendicular, we know that the dot product should be equal to zero. This means we can set the left side of the equation to zero:

0 = |L||L - αS| cos(theta)

Now, we can solve for α by rearranging the equation:

|L - αS| = 0 / |L| cos(theta)

|L - αS| = 0

This means that the magnitude of L - αS must be equal to zero in order for the vectors to be perpendicular. We can rewrite this equation using the Pythagorean theorem:

|L|^2 + |αS|^2 = 0

Since the magnitude of a vector cannot be negative, we know that |αS| must be equal to zero. This means that α must be equal to zero as well.

Therefore, the vector L - αS will be perpendicular to L if α = 0.

I hope this helps! Let me know if you have any further questions.