This problem aims to determine whether the given **vectors** $u$ and $v$ are **parallel** or **not.**

The concept required to solve this problem includes **vector multiplication** like the **cross** and **dot products** and the **angle** between them.Â The **dot product** or commonly known as the **scalar product** of **two vectors** $u$ and $v$ having **magnitude** $|u|$ and $|v|$ can be written as:

\[ u\cdot v = |u||v| \cos \theta \]

Where $\theta$ denotes the **angle** between the **vectors** $u$ and $v$, and $|u|$ and $|v|$ denotes the **magnitude,** whereas \cos\theta represents the **cosine** between the **vectors** $u$ and $v$.

## Expert Answer

To determine the **vectors** $u$ and $v$ as **parallel** or **orthogonal,** we will use the **dot product,** that is:

The **vectors** are **orthogonal** if the angle between them is $90^{\circ}$, or they are **perpendicular**

\[ u\cdot v = 0 \]

But the **vectors** will be **parallel** if they point in the **same** or **opposite direction,** and they never **intersect** each other.

So we have **vectors:**

\[u = <6, 4>;\space v = <-9, 8> \]

Weâ€™ll calculate the **dot product** of the **vectors** to witness whether they are **orthogonal:**

\[u\cdot v=(6)(-9) + (4)(8) \]

\[u\cdot v=-54 + 32 \]

\[u\cdot v=-18 \]

Since the **dot product** does not equal to $0$, we can conclude that $u = <6, 4>$ andÂ $v = <-9, 8>$ are not **orthogonal.**

Now to see if they are **parallel** or not, we will find the **angle** between the given **vectors.** For this, we have to first compute the **magnitude** of $u$ and $v$. The formula to calculate the **magnitude** of a **vector** is given:

\[|u|=\sqrt {x^2 + y^2}\]

For the **magnitude** of $u$:

\[|u|=\sqrt {6^2 + 4^2}\]

\[|u|=\sqrt {36+ 16}\]

\[|u|=\sqrt {52}\]

For the **magnitude** of $v$:

\[|v|=\sqrt {(-9)^2 + 8^2}\]

\[|v|=\sqrt {81+ 64} \]

\[|v|=\sqrt {145} \]

Now to calculate the **angle** between them, we will use the following **equation:**

\[u\cdot v = |u||v| \cos \theta \]

\[\theta=\cos^{-1} (\dfrac{u\cdot v}{|u||v|}) \]

\[\theta=\cos^{-1} (\dfrac{-18}{\sqrt {52} \sqrt {145}}) \]

\[\theta=\cos^{-1} (\dfrac{-18}{86.83}) \]

\[\theta=\cos^{-1} (-0.2077) \]

\[\theta= 101.98^{\circ}\]

Since the **angle** is neither $0$ nor $\pi$, then the **vectors** are **neither parallel nor orthogonal.**

## Numerical Result

The **vectors** $u = <6, 4>$ andÂ $v = <-9, 8>$ are **neither parallel nor**Â **orthogonal.**

## Example

Determine whether the **vectors**Â $u = <3, 15>$ andÂ $v = <-1, 5>$ are **orthogonal** or **parallel** or **neither.**

Computing the **dot product:**

\[u\cdot v=(3)(-1) + (15)(5) \]

\[u\cdot v=-3 + 75 \]

\[u\cdot v=72 \]

So they are not **orthogonal;** we understand this because the **dot-product** of **orthogonal vectors** is equal to **zero.**

Determine if the **two**Â **vectors** are **parallel** by computing the **angle.**

For this, compute the **magnitude** of $u$ and $v$:

\[ |u| = \sqrt {3^2 + 15^2} = \sqrt {234}\]

\[|v|=\sqrt {(-1)^2 + 5^2} = \sqrt {26}\]

Now to calculate the **angle** between them:

\[\theta=\cos^{-1} (\dfrac{72}{\sqrt {234} \sqrt {26}}) \]

\[\theta=22.6^{\circ}\]

If the vectors were **parallel,** their **angle** would be $0$ or $\pi$, so they are **neither parallel** nor **orthogonal.**