Answer
a) Neither
b) Orthogonal
c) Orthogonal
d) Parallel
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) $\vec{a}=\left \langle -5, 3, 7 \right \rangle$ and $\vec{b}=\left \langle 6, -8, 2 \right \rangle$
First, calculate $\vec{a} \cdot \vec{b}$
$\vec{a} \cdot \vec{b}=(-5)(6)+(3)(-8)+(7)(2)=20$, so it's not orthogonal and it's not parallel either, because there is no relationship between the vectors.
To verify, we can calculate $\theta$:
$|\vec{a}|=\sqrt{(-5)^{2}+(3)^{2}+(7)^{2}}=\sqrt{83}$
$|\vec{b}|=\sqrt{(6)^{2}+(-8)^{2}+(2)^{2}}=\sqrt{104}$
$\theta=\arccos {\frac{\vec{a}\cdot \vec{b}}{| \vec{a}| |\vec{b}|}} = \arccos {\frac{20}{\sqrt{83}\sqrt{104}}}=77.57°$
$\therefore$ Answer: Neither
b) $\vec{a}=\left \langle 4,6 \right \rangle$ and $\vec{b}=\left \langle -3,2 \right \rangle$
First, calculate $\vec{a} \cdot \vec{b}$
$\vec{a} \cdot \vec{b}=(4)(-3)+(6)(2)=0$
$\therefore$ Answer: It's orthogonal
c) $\vec{a}=\mathcal{i+2j+5k} $ and $\vec{b}=\mathcal{3i+4j-k} $
First, calculate $\vec{a} \cdot \vec{b}$
$\vec{a} \cdot \vec{b}=(-1)(3)+(2)(4)+(5)(-1)=0$
$\therefore$ Answer: It's orthogonal
d) $\vec{a}=\mathcal{2i+6j-4k} $ and $\vec{b}=\mathcal{-3i-9j+6k} $
First, calculate $\vec{a} \cdot \vec{b}$
$\vec{a} \cdot \vec{b}=(2)(-3)+(6)(-9)+(-4)(6)=-84$, so it's not orthogonal. Parallel: $\vec{a}=-\frac{2}{3}\vec{b}$
$\therefore$ Answer: It's parallel
This can also be verified by calculating $\theta$:
$|\vec{a}|=\sqrt{(2)^{2}+(-4)^{2}+(6)^{2}}=\sqrt{56}$
$|\vec{b}|=\sqrt{(-3)^{2}+(-9)^{2}+(6)^{2}}=\sqrt{126}$
$\theta=\arccos {\frac{\vec{a}\cdot \vec{b}}{| \vec{a}| |\vec{b}|}} = \arccos {\frac{-84}{\sqrt{56}\sqrt{126}}}=180°$
