Let's solve the given problem step by step.

We are given the following:

• $u_1, u_2, u_3$ form an orthonormal basis for an inner product space $U$.
• $u = u_1 + 2u_2 + 3u_3$
• $v = u_1 - 7u_3$

We need to compute:

(a) $\langle u, v \rangle$ (The inner product of $u$ and $v$)

Since $u_1, u_2, u_3$ form an orthonormal basis, we know that the inner product $\langle u_i, u_j \rangle = 0$ for $i \neq j$ and $\langle u_i, u_i \rangle = 1$. Using this, we compute the inner product of $u$ and $v$:

$u = u_1 + 2u_2 + 3u_3$ $v = u_1 - 7u_3$

Now,

$\langle u, v \rangle = \langle u_1 + 2u_2 + 3u_3, u_1 - 7u_3 \rangle$ $= \langle u_1, u_1 \rangle + 2\langle u_2, u_1 \rangle + 3\langle u_3, u_1 \rangle - 7\langle u_1, u_3 \rangle - 14\langle u_2, u_3 \rangle - 21\langle u_3, u_3 \rangle$

Using the properties of the orthonormal basis:

$= 1 + 0 + 0 - 0 - 0 - 21$

Thus,

$\langle u, v \rangle = 1 - 21 = -20$

(b) $\|u\|$ (The norm of $u$)

The norm of $u$ is given by $\|u\| = \sqrt{\langle u, u \rangle}$. We compute:

$\langle u, u \rangle = \langle u_1 + 2u_2 + 3u_3, u_1 + 2u_2 + 3u_3 \rangle$ $= \langle u_1, u_1 \rangle + 4\langle u_2, u_2 \rangle + 9\langle u_3, u_3 \rangle$

Since $\langle u_1, u_1 \rangle = 1$, $\langle u_2, u_2 \rangle = 1$, and $\langle u_3, u_3 \rangle = 1$, we have:

$\langle u, u \rangle = 1 + 4 + 9 = 14$

Therefore,

$\|u\| = \sqrt{14}$

(c) The projection $u^\parallel$ of $u$ onto $v$

The projection of $u$ onto $v$ is given by:

$u^\parallel = \frac{\langle u, v \rangle}{\langle v, v \rangle} v$

We already know that $\langle u, v \rangle = -20$. Now we compute $\langle v, v \rangle$:

$\langle v, v \rangle = \langle u_1 - 7u_3, u_1 - 7u_3 \rangle$ $= \langle u_1, u_1 \rangle + 49\langle u_3, u_3 \rangle = 1 + 49 = 50$

Thus,

$u^\parallel = \frac{-20}{50} v = -\frac{2}{5} v = -\frac{2}{5}(u_1 - 7u_3)$

This simplifies to:

$u^\parallel = -\frac{2}{5}u_1 + \frac{14}{5}u_3$

(d) The angle $\theta$ between $u$ and $v$

The cosine of the angle $\theta$ between $u$ and $v$ is given by:

$\cos \theta = \frac{\langle u, v \rangle}{\|u\| \|v\|}$

We already know:

$\langle u, v \rangle = -20, \quad \|u\| = \sqrt{14}, \quad \|v\| = \sqrt{\langle v, v \rangle} = \sqrt{50}$

Thus,

$\cos \theta = \frac{-20}{\sqrt{14} \cdot \sqrt{50}} = \frac{-20}{\sqrt{700}} = \frac{-20}{10\sqrt{7}} = \frac{-2}{\sqrt{7}}$

Therefore, the angle $\theta$ is:

$\theta = \cos^{-1}\left( \frac{-2}{\sqrt{7}} \right)$

This gives the angle in terms of the arccosine function.

Would you like further details or explanations? Here are 5 related questions:

1. What are the properties of an orthonormal basis in an inner product space?
2. How is the projection of a vector onto another vector used in applications?
3. What happens if the angle $\theta$ between two vectors is $0^\circ$ or $90^\circ$?
4. How can we generalize the inner product to other dimensions or different inner product spaces?
5. How is the cosine of the angle between two vectors used in machine learning models?

Tip: Always check whether your vectors form an orthonormal basis before applying standard inner product formulas, as it simplifies computations considerably.