1. ## Unit Vectors

I have to find two unit vectors that are parallel to

<3, 1, 2>

And then I have to write each vector as the product of its magnitude and a unit vector.

--------

My approach:

We know that two vectors are parallel if their cross product is equal to 0.

$\mathbf{a} \times \mathbf{b} = \mathbf{A}_{\times} \mathbf{b} = \begin{bmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3 \end{bmatrix}$

Using the above, I plugged in a_1, b_1, c_1 and then augmented it with the 0 vector, but I just ended up with a_1 = 0, b_1 = 0, and c_1 = 0 which wasn't very helpful. Any ideas?

*EDIT* I guess Wikipedia math code doesn't conform to this site;

It's mean to be:

a x b = [[0, -a_3 a_2], [a_3, 0, -a_1], [-a_2, a_1, 0]] multiplied with the vector [b_1, b_2, b_3]; of course there is also the determinant method.

2. Originally Posted by Ideasman
I have to find two unit vectors that are parallel to

<3, 1, 2>

And then I have to write each vector as the product of its magnitude and a unit vector.

--------

My approach:

We know that two vectors are parallel if their cross product is equal to 0.

$\mathbf{a} \times \mathbf{b} = \mathbf{A}_{\times} \mathbf{b} = \begin{bmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3 \end{bmatrix}$

Using the above, I plugged in a_1, b_1, c_1 and then augmented it with the 0 vector, but I just ended up with a_1 = 0, b_1 = 0, and c_1 = 0 which wasn't very helpful. Any ideas?

*EDIT* I guess Wikipedia math code doesn't conform to this site;

It's mean to be:

a x b = [[0, -a_3 a_2], [a_3, 0, -a_1], [-a_2, a_1, 0]] multiplied with the vector [b_1, b_2, b_3]; of course there is also the determinant method.
You're thinking about it too much. Just use any scalar.

That is, two vectors are parallel if they are a scalar of the other. You can check with using cross products to show they are parallel.

So, <6, 2, 4> and <9, 3, 6>. Now find the norm of each and you should be able to take it from there.

3. Originally Posted by AfterShock
You're thinking about it too much. Just use any scalar.

That is, two vectors are parallel if they are a scalar of the other. You can check with using cross products to show they are parallel.

So, <6, 2, 4> and <9, 3, 6>. Now find the norm of each and you should be able to take it from there.
What Aftershock said, but with an addendum:
You need TWO unit vectors along the same line. They will necessarily be in opposite directions. So find one unit vector by taking the norm of the given vector and dividing that into your given vector. (Aftershock's method.) The second unit vector will merely be in the opposite direction of this, so multiply your first vector by a -1.

-Dan

4. Hello, Ideasman!

A strange question . . .

Find two unit vectors that are parallel to: . $\vec{v}\:=\:\langle 3,\,1,\,2\rangle$

The magnitude of $\vec{v}$ is: . $|\vec{v}| \:=\:\sqrt{3^2 + 1^2 + 2^2} \:=\:\sqrt{14}$

The two unit vectors are:

. . $u_1\:=\:\frac{\vec{v}}{|\vec{v}|} \:=\:\frac{\langle 3,\,1,\,2\rangle}{\sqrt{14}}\:=\:\left\langle \frac{3}{\sqrt{14}},\:\frac{1}{\sqrt{14}},\:\frac{ 2}{\sqrt{14}}\right\rangle$

. . $u_2\:=\:\text{-}\frac{\vec{v}}{|\vec{v}|} \:=\:\text{-}\frac{\langle3,\,1,\,2\rangle}{\sqrt{14}} \:=\:\left\langle\frac{-3}{\sqrt{14}},\,\frac{-1}{\sqrt{14}},\,\frac{-2}{\sqrt{14}}\right\rangle
$

Write each vector as the product of its magnitude and a unit vector.
This is the strange part: a unit vector has magnitude 1.
So, what the point of this task? ... multiplying by one?