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 $vec{i}$, $vec{j}$, and $vec{k}$.
(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 $vec{b}\timesvec{a}$.
homework #2
due thursday, sep. 18, 2025
late homework will not be accepted. the homework should be placed on the instructors desk before the start of the class on the due date. there should be some penalties on unstapled homework.

Explanation:

Step1: Define vectors

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

Step2: Compute magnitude of \(\vec{a}\)

The magnitude of a vector \(\vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k}\) is \(|\vec{a}|=\sqrt{a_x^{2}+a_y^{2}+a_z^{2}}\). So \(|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{4 + 1+9}=\sqrt{14}\)

Step3: Compute magnitude of \(\vec{b}\)

For \(\vec{b}=b_x\vec{i}+b_y\vec{j}+b_z\vec{k}\), \(|\vec{b}|=\sqrt{b_x^{2}+b_y^{2}+b_z^{2}}\). So \(|\vec{b}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1 + 4+9}=\sqrt{14}\)

Step4: Compute the dot - product \(\vec{a}\cdot\vec{b}\)

\(\vec{a}\cdot\vec{b}=a_xb_x + a_yb_y+a_zb_z=(2\times1)+(1\times(-2))+(-3\times3)=2-2 - 9=-9\)

Step5: Compute the angle between \(\vec{a}\) and \(\vec{b}\)

The formula for the angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\) is \(\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\). Substituting the values we found: \(\cos\theta=\frac{-9}{\sqrt{14}\times\sqrt{14}}=-\frac{9}{14}\), so \(\theta=\arccos(-\frac{9}{14})\)

Step6: 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 \(\vec{b}\), \(\hat{b}=\frac{\vec{b}}{|\vec{b}|}=\frac{1}{\sqrt{14}}\vec{i}-\frac{2}{\sqrt{14}}\vec{j}+\frac{3}{\sqrt{14}}\vec{k}\)

Step7: Compute the cosine angles of unit vectors

For a unit vector \(\hat{u}=u_x\vec{i}+u_y\vec{j}+u_z\vec{k}\), the cosine of the angles with the axes are \(\cos\alpha = u_x\), \(\cos\beta=u_y\), \(\cos\gamma = u_z\)
For \(\hat{a}\): \(\cos\alpha_a=\frac{2}{\sqrt{14}}\), \(\cos\beta_a=\frac{1}{\sqrt{14}}\), \(\cos\gamma_a=-\frac{3}{\sqrt{14}}\)
For \(\hat{b}\): \(\cos\alpha_b=\frac{1}{\sqrt{14}}\), \(\cos\beta_b=-\frac{2}{\sqrt{14}}\), \(\cos\gamma_b=\frac{3}{\sqrt{14}}\)

Step8: Compute the cross - product \(\vec{a}\times\vec{b}\)

\(\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)\)
\(=\vec{i}(3 - 6)-\vec{j}(6 + 3)+\vec{k}(-4 - 1)=-3\vec{i}-9\vec{j}-5\vec{k}\)
And \(\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}=3\vec{i}+9\vec{j}+5\vec{k}\)

Answer:

Magnitude of \(\vec{a}\): \(\sqrt{14}\), Magnitude of \(\vec{b}\): \(\sqrt{14}\), \(\vec{a}\cdot\vec{b}=-9\), Angle between \(\vec{a}\) and \(\vec{b}\): \(\arccos(-\frac{9}{14})\), 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 \(\vec{b}\): \(\frac{1}{\sqrt{14}}\vec{i}-\frac{2}{\sqrt{14}}\vec{j}+\frac{3}{\sqrt{14}}\vec{k}\), Cosine angles of \(\hat{a}\): \(\cos\alpha_a=\frac{2}{\sqrt{14}}\), \(\cos\beta_a=\frac{1}{\sqrt{14}}\), \(\cos\gamma_a=-\frac{3}{\sqrt{14}}\), Cosine angles of \(\hat{b}\): \(\cos\alpha_b=\frac{1}{\sqrt{14}}\), \(\cos\beta_b=-\frac{2}{\sqrt{14}}\), \(\cos\gamma_b=\frac{3}{\sqrt{14}}\), \(\vec{a}\times\vec{b}=-3\vec{i}-9\vec{j}-5\vec{k}\), \(\vec{b}\times\vec{a}=3\vec{i}+9\vec{j}+5\vec{k}\)