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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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- Thread starter C172Driver
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In summary: Since there are an infinite number of vectors that could satisfy this equation, it would be impossible to list them all. However, the following are two vectors that satisfy the equation: \vec{u} = x\vec{i} + y\vec{j} + \sqrt{1-x^2-y^2}\vec{k} = (-1,0)_3 and \vec{u} = x\vec{i} + y\vec{j} + \sqrt{1-x^2+y^2}\vec{k} = (0,1)_3.

- #1

- 1

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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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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.

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C172Driver said:

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.

The angle between two vectors in 3 dimensions can be calculated using the dot product formula:

θ = cos^-1((a1b1 + a2b2 + a3b3) / (sqrt(a1^2 + a2^2 + a3^2) * sqrt(b1^2 + b2^2 + b3^2)))

where a and b are the components of the two vectors.

A unit vector is a vector with a magnitude of 1. It is often used to represent direction without changing the scale of the vector.

The magnitude of a vector can be found using the Pythagorean theorem:

|a| = sqrt(a1^2 + a2^2 + a3^2)

where a1, a2, and a3 are the components of the vector.

No, the angle between two vectors in 3 dimensions cannot be greater than 180 degrees. This is because the dot product formula takes the inverse cosine of the angle, which has a range of 0 to 180 degrees.

The angle between two vectors is affected by their orientations. If the two vectors are parallel, the angle between them is 0 degrees. If they are antiparallel, the angle is 180 degrees. If they are perpendicular, the angle is 90 degrees. Any other orientation will result in an angle between 0 and 180 degrees.

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