Answer
a) Parallel
b) Neither
c) Orthogonal
Work Step by Step
We must start from the following information:
1. Orthogonal: If $\vec{a} \cdot \vec{b}=0$ or $\theta=90°$
2. Parallel: When one of the vectors multiplied by a scalar is the other vector, either $\theta=0$ or $\theta=180°$
3. Neither: When the above conditions are not met. $\theta$ can be calculated and must be different from $90°, 0°$, or $180°$
Now, if we want to calculate $\theta$, we have the formula: $\theta=arccos\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
a) $\vec{u}=\left \langle -3, 9, 6 \right \rangle$ and $\vec{v}=\left \langle 4, -12, -8 \right \rangle$
First calculate $\vec{u} \cdot \vec{v}$:
$\vec{u} \cdot \vec{v}=(-3)(4)+(9)(-12)+(6)(-8)=-168$, so they aren't orthogonal.
To know if it's parallel, we need to find a relationship between them:
Using the x-coordinate we have:
$-3x=4$ solving for x $x=\frac{-3}{4}$
If we multiply each component of the second vector by the obtained value of x, we must obtain the first vector:
x-component: $4(\frac{-3}{4})=-3$
y-component: $-12(\frac{-3}{4})=9$
z-component: $-8(\frac{-3}{4})=6$
Since we do get the first vector, it means: $\vec{u}=\frac{-3}{4}\vec{v}$
$\therefore$ Are parallel
b) $\vec{u}=\mathcal{i-j+2k}$ and $\vec{v}=\mathcal{2i-j+k}$
First calculate $\vec{u} \cdot \vec{v}$:
$\vec{u} \cdot \vec{v}=(1)(2)+(-1)(-1)+(2)(1)=5$, so they aren't orthogonal.
To know if it's parallel, we need to find a relationship between them:
Using the i-coordinate we have:
$2x=1$ solving for x $x=\frac{1}{2}$
If we multiply each component of the second vector by the obtained value of x, we must obtain the first vector:
i-component: $2(\frac{1}{2})=1$
j-component: $-1(\frac{1}{2})=\frac{-1}{2}$
k-component: $1(\frac{1}{2})=\frac{1}{2}$
Since multiplying does not give us the first vector, it means that they are not parallel.
$\therefore$ Neither
We can verify by calculating $\theta$
c) $\vec{u}=\left \langle a, b, c \right \rangle$ and $\vec{v}=\left \langle -b, a, 0 \right \rangle$
First calculate $\vec{u} \cdot \vec{v}$:
$\vec{u} \cdot \vec{v}=(a)(-b)+(a)(b)+(c)(0)=-ab+ab=0$
$\therefore$ Are orthogonal
