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Angle between a vector and a unit vectors with 3 dimensions.

  • Thread starter C172Driver
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  • #1
Find a two unit vectors that make the angle [itex]\pi[/itex]/3 with the vector [itex]\vec{v}[/itex] = [itex]\vec{i}[/itex] + 2[itex]\vec{j}[/itex] + 3[itex]\vec{k}[/itex]. "That isn't asking much since there are apparently infinite such vectors" - Prof.

I get as far as to say that [itex]\pi[/itex]/3 = arccos( 1/2 ) and that [itex]\frac{v \bullet w}{|v||w|}[/itex] = 1/2 but from there I am lost as to how to figure this out.

Help is appreciated! Thanks!
 

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  • #2
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I am not sure if you are familiar with linear algebra but it may prove useful. Let us assume that the vector only rotates in the xy plane (seems fair enough).

Well, if we wanted to rotate the unit vector i by sixty degrees we can describe it by a matrix operation such that : A*i= <1/2, √3/2>.

That new vector is what i would be rotated by sixty degrees.

Once you find the 2x2 matrix A you can multiply that by the x,y components of v and get the new rotated angle.

Note: z stays the same.
 
  • #3
Curious3141
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Find a two unit vectors that make the angle [itex]\pi[/itex]/3 with the vector [itex]\vec{v}[/itex] = [itex]\vec{i}[/itex] + 2[itex]\vec{j}[/itex] + 3[itex]\vec{k}[/itex]. "That isn't asking much since there are apparently infinite such vectors" - Prof.

I get as far as to say that [itex]\pi[/itex]/3 = arccos( 1/2 ) and that [itex]\frac{v \bullet w}{|v||w|}[/itex] = 1/2 but from there I am lost as to how to figure this out.

Help is appreciated! Thanks!
Let the required unit vector be denoted by [itex]\vec{u} = x\vec{i} + y\vec{j} + \sqrt{1-x^2-y^2}\vec{k}[/itex]. The reason for this form is so that its norm equals one (verify this).

Now ##\vec{u}.\vec{v}= \frac{1}{2}\sqrt{14}##, yes?

Just do the algebra. All you need to do is find 2 (of the infinitude of) possible values for ##\vec{u}##. To make your life easier, put ##x=0## and ##y=0## in turn, and solve each resulting quadratic.
 

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