Finding a unit vector perpendicular to both a and b using the dot product?

rl127

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11 years ago

Finding a unit vector perpendicular to both a and b using the dot product?

Hi there
i really need help with this vector question.how do i find a vector that is perpendicular to both a[1,2,3] and b[2,-2,-6] using the dot product. Please help me. I would really appreciate it. thank you Slight Smile

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Smoren

wrote...

11 years ago

__ i __ 1 __ 2
__ j __ 2 __ -2
__ k _ 3 __ -6

n= -12i -2k +6j -4k +6i +6j = -6i +12j -6k
divided by -2
n`=[1 -2 1]
|n|=root(1² + 2² + 1²) = root6
unit vector = [1 -2 1] / root6

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smokinjoe531

wrote...

11 years ago

Let c be the vector (x, y, z).

The dot product of perpendicular vectors is equal to zero. Find a dot c and b dot c and set them equal to zero. The magnitude of c needs to be one so ?(x^2+y^2+z^2) = 1. Or more simply x^2 + y^2 + z^2 = 1.

This will give you a system of three equations with three unknowns. Solve the system to get your answer.

P.S. It would be a lot easier to use the cross product.

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LESHA

wrote...

11 years ago

Dot product? Are you sure it isn't cross product? Cross product is much easier.

With the dot product, we know that two vectors, A and B, if their dot product is zero, then the vectors are perpendicular.

A = [1,2,3]
B = [2,-2,-6]
C = [x,y,z], we need to find this one

A?C = x+2y+3z=0
B?C = 2x-2y-6z=0

Since you need the vector to be a unit vector, we have a 3rd equation:
x^2+y^2+z^2=1^2, or
x^2+y^2+z^2=1

Using www.wolframalpha.com (didn't feel like doing this complex solving), I got:

x = -1/sqrt(6)
y = sqrt(2/3)
z = 1/sqrt(6)

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