Question

1. consider 2 points a (2,1, - 3) and b (1, - 2,3) for vectors $vec{a}$ and $vec{b}$, respectively, in a 3 - dimensional cartesian coordinate.

(a) find vectors $vec{a}$ and $vec{b}$ representing points a and b, using the base vectors $hat{i}$, $hat{j}$, and $hat{k}$.
(b) compute the magnitude of each vector.
(c) compute the unit vectors for each vector.
(d) find cosine angles ($alpha,\beta$, and $gamma$) of each vector.
(e) compute $vec{a}cdotvec{b}$ and $vec{b}cdotvec{a}$.
(f) compute the angle between $vec{a}$ and $vec{b}$.
(g) compute $vec{a}\timesvec{b}$ and $vec{b}\timesvec{a}$.

Explanation:

Step1: Find vectors $\vec{a}$ and $\vec{b}$

Given points $a(2,1, - 3)$ and $b(1,-2,3)$, then $\vec{a}=2\vec{i}+1\vec{j}-3\vec{k}$ and $\vec{b}=1\vec{i}-2\vec{j}+3\vec{k}$.

Step2: Compute the magnitude of each vector

The magnitude of $\vec{a}$, $|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{4 + 1+9}=\sqrt{14}$.
The magnitude of $\vec{b}$, $|\vec{b}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1 + 4+9}=\sqrt{14}$.

Step3: Compute the unit - vectors

The unit - vector of $\vec{a}$, $\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{2}{\sqrt{14}}\vec{i}+\frac{1}{\sqrt{14}}\vec{j}-\frac{3}{\sqrt{14}}\vec{k}$.
The unit - vector of $\vec{b}$, $\hat{b}=\frac{\vec{b}}{|\vec{b}|}=\frac{1}{\sqrt{14}}\vec{i}-\frac{2}{\sqrt{14}}\vec{j}+\frac{3}{\sqrt{14}}\vec{k}$.

Step4: Compute the dot - products

$\vec{a}\cdot\vec{b}=(2\vec{i}+1\vec{j}-3\vec{k})\cdot(1\vec{i}-2\vec{j}+3\vec{k})=2\times1+1\times(-2)+(-3)\times3=2 - 2-9=-9$.
$\vec{b}\cdot\vec{a}=-9$ (since $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$).

Step5: Find the cosine angles

The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}=\frac{-9}{\sqrt{14}\times\sqrt{14}}=-\frac{9}{14}$.
Let $\vec{a}=a_{x}\vec{i}+a_{y}\vec{j}+a_{z}\vec{k}$ and $\vec{b}=b_{x}\vec{i}+b_{y}\vec{j}+b_{z}\vec{k}$.
The cosine of the angle $\alpha$ between $\vec{a}$ and $\vec{i}$ is $\cos\alpha=\frac{\vec{a}\cdot\vec{i}}{|\vec{a}|}=\frac{2}{\sqrt{14}}$, for $\vec{b}$, $\cos\alpha_{b}=\frac{\vec{b}\cdot\vec{i}}{|\vec{b}|}=\frac{1}{\sqrt{14}}$.
The cosine of the angle $\beta$ between $\vec{a}$ and $\vec{j}$ is $\cos\beta=\frac{\vec{a}\cdot\vec{j}}{|\vec{a}|}=\frac{1}{\sqrt{14}}$, for $\vec{b}$, $\cos\beta_{b}=\frac{\vec{b}\cdot\vec{j}}{|\vec{b}|}=-\frac{2}{\sqrt{14}}$.
The cosine of the angle $\gamma$ between $\vec{a}$ and $\vec{k}$ is $\cos\gamma=\frac{\vec{a}\cdot\vec{k}}{|\vec{a}|}=-\frac{3}{\sqrt{14}}$, for $\vec{b}$, $\cos\gamma_{b}=\frac{\vec{b}\cdot\vec{k}}{|\vec{b}|}=\frac{3}{\sqrt{14}}$.

Step6: Compute the angle between $\vec{a}$ and $\vec{b}$

$\theta=\arccos(-\frac{9}{14})\approx130.3^{\circ}$.

Step7: Compute the cross - products

$\vec{a}\times\vec{b}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\2&1&-3\\1&-2&3\end{vmatrix}$$

=\vec{i}(1\times3-(-2)\times(-3))-\vec{j}(2\times3 - 1\times(-3))+\vec{k}(2\times(-2)-1\times1)=\vec{i}(3 - 6)-\vec{j}(6 + 3)+\vec{k}(-4 - 1)=-3\vec{i}-9\vec{j}-5\vec{k}$.
$\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}=3\vec{i}+9\vec{j}+5\vec{k}$.

Answer:

• Magnitude of $\vec{a}$: $\sqrt{14}$, Magnitude of $\vec{b}$: $\sqrt{14}$
• Unit - vector of $\vec{a}$: $\frac{2}{\sqrt{14}}\vec{i}+\frac{1}{\sqrt{14}}\vec{j}-\frac{3}{\sqrt{14}}\vec{k}$, Unit - vector of $\vec{b}$: $\frac{1}{\sqrt{14}}\vec{i}-\frac{2}{\sqrt{14}}\vec{j}+\frac{3}{\sqrt{14}}\vec{k}$
• $\vec{a}\cdot\vec{b}=-9$, $\vec{b}\cdot\vec{a}=-9$
• Cosine of the angle between $\vec{a}$ and $\vec{b}$: $-\frac{9}{14}$, Angle between $\vec{a}$ and $\vec{b}\approx130.3^{\circ}$
• Cosine of angles with axes for $\vec{a}$: $\cos\alpha=\frac{2}{\sqrt{14}},\cos\beta=\frac{1}{\sqrt{14}},\cos\gamma =-\frac{3}{\sqrt{14}}$; for $\vec{b}$: $\cos\alpha_{b}=\frac{1}{\sqrt{14}},\cos\beta_{b}=-\frac{2}{\sqrt{14}},\cos\gamma_{b}=\frac{3}{\sqrt{14}}$
• $\vec{a}\times\vec{b}=-3\vec{i}-9\vec{j}-5\vec{k}$, $\vec{b}\times\vec{a}=3\vec{i}+9\vec{j}+5\vec{k}$