- #1

- 1

- 0

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!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter C172Driver
- Start date

- #1

- 1

- 0

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!

- #2

- 92

- 8

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

Homework Helper

- 2,843

- 87

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.

Share:

- Replies
- 11

- Views
- 5K