# Determine whether the given vectors are orthogonal, parallel, or neither. u = ⟨6, 4⟩, v = ⟨-9, 8⟩

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. 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 $v$.

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:

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

We’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 $u = <6, 4>$ and  $v = <-9, 8>$ 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 $u = <6, 4>$ and  $v = <-9, 8>$ are neither parallel nor orthogonal.

## Example

Determine whether the vectors $u = <3, 15>$ and  $v = <-1, 5>$ 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 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.