Section 0 — Mathematical Foundations
1. Vector Algebra
We are given two vectors in \(\mathbb{R}^3\):
\[
\vec{a} = [2,\,1,\,-3], \qquad \vec{b} = [4,\,-2,\,1]
\]
1) Magnitude (length) of a vector
For \(\vec{v} = [v_x, v_y, v_z]\),
\[
|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}
\]
2) Dot product
For \(\vec{a} = [a_x,a_y,a_z]\) and \(\vec{b} = [b_x,b_y,b_z]\),
\[
\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z
\]
3) Cross product
For \(\vec{a} = [a_x,a_y,a_z]\) and \(\vec{b} = [b_x,b_y,b_z]\),
\[
\vec{a}\times\vec{b} =
\begin{bmatrix}
a_y b_z - a_z b_y \\
a_z b_x - a_x b_z \\
a_x b_y - a_y b_x
\end{bmatrix}
\]
4) Angle between two vectors
If \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\), then
\[
\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}| \cos\theta
\]
So,
\[
\theta = \arccos\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)
\]
(a) Magnitude of each vector
Magnitude of \(\vec{a}\)
Using \(|\vec{a}| = \sqrt{2^2 + 1^2 + (-3)^2}\):
\[
|\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}
\]
Magnitude of \(\vec{b}\)
Using \(|\vec{b}| = \sqrt{4^2 + (-2)^2 + 1^2}\):
\[
|\vec{b}| = \sqrt{16 + 4 + 1} = \sqrt{21}
\]
Answer (a):
\[
|\vec{a}|=\sqrt{14}, \qquad |\vec{b}|=\sqrt{21}
\]
(b) Dot product \(\vec{a}\cdot\vec{b}\)
Compute component-by-component:
\[
\vec{a}\cdot\vec{b} = (2)(4) + (1)(-2) + (-3)(1)
\]
\[
\vec{a}\cdot\vec{b} = 8 - 2 - 3 = 3
\]
Answer (b):
\[
\vec{a}\cdot\vec{b} = 3
\]
(c) Cross product \(\vec{a}\times\vec{b}\)
Let \(\vec{a}=[2,1,-3]\) and \(\vec{b}=[4,-2,1]\).
\(x\)-component
\[
(\vec{a}\times\vec{b})_x = a_y b_z - a_z b_y = (1)(1) - (-3)(-2) = 1 - 6 = -5
\]
\(y\)-component
\[
(\vec{a}\times\vec{b})_y = a_z b_x - a_x b_z = (-3)(4) - (2)(1) = -12 - 2 = -14
\]
\(z\)-component
\[
(\vec{a}\times\vec{b})_z = a_x b_y - a_y b_x = (2)(-2) - (1)(4) = -4 - 4 = -8
\]
So,
\[
\vec{a}\times\vec{b} = [-5,\,-14,\,-8]
\]
Answer (c):
\[
\vec{a}\times\vec{b} = [-5,\,-14,\,-8]
\]
(d) Angle between \(\vec{a}\) and \(\vec{b}\)
Use
\[
\theta = \arccos\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)
\]
We already found:
- \(\vec{a}\cdot\vec{b} = 3\)
- \(|\vec{a}| = \sqrt{14}\)
- \(|\vec{b}| = \sqrt{21}\)
Substitute:
\[
\theta = \arccos\!\left(\frac{3}{\sqrt{14}\sqrt{21}}\right)
\]
Combine the radicals:
\[
\sqrt{14}\sqrt{21}=\sqrt{294}
\]
So the exact form is:
\[
\theta = \arccos\!\left(\frac{3}{\sqrt{294}}\right)
\]
Optional numeric approximation:
- \(\sqrt{294}\approx 17.146\)
- \(\frac{3}{\sqrt{294}}\approx 0.175\)
So,
\[
\theta \approx \arccos(0.175)\approx 1.395\text{ rad}\approx 79.9^\circ
\]
Answer (d):
\[
\theta = \arccos\!\left(\frac{3}{\sqrt{294}}\right)\approx 79.9^\circ
\]
Final answers (summary)
\[
|\vec{a}|=\sqrt{14}, \qquad |\vec{b}|=\sqrt{21}
\]
\[
\vec{a}\cdot\vec{b}=3
\]
\[
\vec{a}\times\vec{b}=[-5,\,-14,\,-8]
\]
\[
\theta=\arccos\!\left(\frac{3}{\sqrt{294}}\right)\approx 79.9^\circ
\]