Find two unit vectors that make a 30 degree angle with v = <3,4>

To find two unit vectors that make a 30-degree angle with the vector **v** = \(\langle 3, 4 \rangle\), we can use the dot product formula. The dot product between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by:

\[
\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos \theta
\]

Where:
- \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) are the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\),
- \(\theta\) is the angle between the vectors.

Since we are looking for unit vectors \(\mathbf{u}\), \(\|\mathbf{u}\|\) is 1. Also, \(\theta = 30^\circ\).

The dot product formula simplifies to:

\[
\mathbf{u} \cdot \mathbf{v} = \|\mathbf{v}\| \cos 30^\circ
\]

The magnitude of \(\mathbf{v} = \langle 3, 4 \rangle\) is:

\[
\|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]

So, the dot product becomes:

\[
\mathbf{u} \cdot \mathbf{v} = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}
\]

Now, let's assume the unit vector \(\mathbf{u} = \langle u_1, u_2 \rangle\). The dot product \(\mathbf{u} \cdot \mathbf{v}\) is:

\[
u_1 \cdot 3 + u_2 \cdot 4 = \frac{5\sqrt{3}}{2}
\]

Also, since \(\mathbf{u}\) is a unit vector, we have:

\[
u_1^2 + u_2^2 = 1
\]

We now have the system of equations:

\[
3u_1 + 4u_2 = \frac{5\sqrt{3}}{2}
\]
\[
u_1^2 + u_2^2 = 1
\]

We can solve this system to find the two unit vectors.

### Solving the System

1. Express \( u_2 \) from the first equation:

\[
u_2 = \frac{\frac{5\sqrt{3}}{2} - 3u_1}{4}
\]

2. Substitute \( u_2 \) in the second equation:

\[
u_1^2 + \left(\frac{\frac{5\sqrt{3}}{2} - 3u_1}{4}\right)^2 = 1
\]

Simplifying the equation will yield a quadratic in \( u_1 \). Solving this quadratic will give us two possible values for \( u_1 \), and corresponding \( u_2 \) values. These will be the components of the two unit vectors.

Let me solve this algebraically for you.The two unit vectors that make a 30-degree angle with the vector \(\mathbf{v} = \langle 3, 4 \rangle\) are:

1. \(\mathbf{u}_1 = \left\langle -\frac{2}{5} + \frac{3\sqrt{3}}{10}, \frac{3}{10} + \frac{2\sqrt{3}}{5} \right\rangle\)
2. \(\mathbf{u}_2 = \left\langle \frac{2}{5} + \frac{3\sqrt{3}}{10}, -\frac{3}{10} + \frac{2\sqrt{3}}{5} \right\rangle\)

These vectors satisfy the condition of having a 30-degree angle with \(\mathbf{v}\) and are both unit vectors.

Would you like further details on how these values were derived, or do you have any other questions?

Here are five related questions you might find interesting:
1. How do you calculate the angle between two general vectors?
2. What is the significance of the dot product in vector analysis?
3. How do you find the projection of one vector onto another?
4. What are the properties of unit vectors?
5. How do vectors behave under linear transformations?

**Tip:** In vector calculus, understanding how angles between vectors relate to their dot product can greatly simplify the analysis of geometric problems.

Learn how to find two unit vectors that make a 30-degree angle with the vector v = <3, 4>. Explore the dot product formula and trigonometric principles applied to vectors. Suitable for advanced high school students and beyond.

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