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Vector Operations

Vector operations include addition, scalar multiplication, magnitude, dot product, and cross product.

$$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$$

Angle Between Vectors

Find the angle between $\vec{a}=\langle 2,-1,2\rangle$ and $\vec{b}=\langle 1,2,2\rangle$.

$$\vec{a}\cdot\vec{b}=2(1)+(-1)(2)+2(2)=4$$
$$|\vec{a}|=3, \qquad |\vec{b}|=3$$
$$\theta=\cos^{-1}\left(\frac{4}{9}\right)=63.6^\circ$$

Final answer: $63.6^\circ$.

Vector Magnitude and Unit Vector — CE Board

Find the magnitude and unit vector of $\vec{v} = \langle 3,\,-4,\,0 \rangle$.

Magnitude:

$$|\vec{v}| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Unit vector (divide each component by the magnitude):

$$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{5}\langle 3,\,-4,\,0\rangle = \left\langle\frac{3}{5},\,-\frac{4}{5},\,0\right\rangle$$

Verify: $|\hat{v}| = \sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = 1$ ✓

Cross Product and Parallelogram Area — CE Board

Given $\vec{a} = \langle 2,\,1,\,-1\rangle$ and $\vec{b} = \langle 1,\,-1,\,2\rangle$, find $\vec{a}\times\vec{b}$ and the area of the parallelogram they form.

Expand the determinant:

$$\vec{a}\times\vec{b} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\2&1&-1\\1&-1&2\end{vmatrix}$$
$$= \vec{i}[(1)(2)-(-1)(-1)] - \vec{j}[(2)(2)-(-1)(1)] + \vec{k}[(2)(-1)-(1)(1)]$$
$$= \vec{i}(2-1) - \vec{j}(4+1) + \vec{k}(-2-1) = \langle 1,\,-5,\,-3\rangle$$

Area of parallelogram = magnitude of cross product:

$$\text{Area} = |\vec{a}\times\vec{b}| = \sqrt{1^2+5^2+3^2} = \sqrt{35} \approx 5.92 \text{ sq. units}$$

Problem: Angle from Dot Product

Find the angle between A = 3i + 4j and B = 5i - 2j.

$$A\cdot B=3(5)+4(-2)=7$$
$$|A|=5,\quad |B|=\sqrt{29}$$
$$\cos\theta=\frac{7}{5\sqrt{29}}\Rightarrow \theta=74.9^\circ$$

Answer: The angle is about 74.9°.

Problem: Unit Vector

Find the unit vector in the direction of 6i - 8j.

$$|v|=\sqrt{6^2+(-8)^2}=10$$

Answer: The unit vector is $0.6i-0.8j$.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w240

MSTE - Physics/Dynamics / Vectors / MSTE April 2025

The cross-product $A\times B = 0$ and the dot-product $A\cdot B = 30$. If $A = 4i + 2j$, find B.

  1. $6i - 3j$
  2. $-6i + 3j$
  3. $-6i - 3j$
  4. $6i + 3j$
Let $B = B_x i + B_y j$.

From the cross product (for 2D vectors in the xy-plane, $A\times B = (A_xB_y - B_xA_y)k$):
$(4i + 2j)\times(B_x i + B_y j) = 0$
$4B_y - 2B_x = 0$
$-2B_x + 4B_y = 0$   (1)

From the dot product:
$(4i + 2j)\cdot(B_x i + B_y j) = 30$
$4B_x + 2B_y = 30$   (2)

Solving (1) and (2) simultaneously:
$B_x = 6$, $B_y = 3$
$\boxed{B = 6i + 3j}$

Note: $A\times B = 0$ means the vectors are parallel, so B is a positive multiple of A.

Question Bank: w245

MSTE - Physics/Dynamics / Vectors / MSTE April 2025

Given the following vectors: $A = 6i - 2j - k$, $B = 2i + 5j + 2k$, $P = 2i - 4j + k$, and $Q = -6i + 12j - 3k$. Which statement is correct?

  1. A & B are parallel; P & Q are parallel
  2. A & B are orthogonal; P & Q are orthogonal
  3. A & B are parallel; P & Q are orthogonal
  4. A & B are orthogonal; P & Q are parallel
Two vectors are orthogonal if their dot product is zero.
$A\cdot B = 6(2) - 2(5) - 1(2) = 12 - 10 - 2 = 0$
Therefore A and B are orthogonal.

Two vectors are parallel if their cross product is zero, or by inspection if one is a scalar multiple of the other: $U = \pm cV$ where $c$ is a constant.
$Q = -6i + 12j - 3k = -3(2i - 4j + k) = -3P$
Therefore P and Q are parallel.

$\boxed{\text{A \& B are orthogonal; P \& Q are parallel}}$

Question Bank: w275

MSTE - Physics/Dynamics / Vectors / MSTE April 2025

Find the volume of the parallelepiped whose edges are represented by the following vectors: $A = 2i - 3j + 4k$, $B = i + 2j - k$, and $C = 3i - j + 2k$.

  1. 5
  2. 6
  3. 7
  4. 8
The volume of a parallelepiped is the absolute value of the scalar triple product:
$V = A\cdot(B\times C)$

Evaluate as a 3×3 determinant of the components:
$V = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = -7$

Volume is a magnitude, so take the absolute value:
$\boxed{V = 7\ \text{unit}^{3}}$