1. Scalar (or Dot) Product of Two Vectors
The scalar (or dot) product of two vectors $\vec{u}$ and $\vec{v}$ is a scalar quantity defined geometrically by:
$$ \vec{u} \cdot \vec{v} = || \vec{u} || \, || \vec{v} || \cos \theta $$
where $|| \vec{u} ||$ and $|| \vec{v} ||$ are the magnitudes of the vectors, and $\theta$ is the angle between them.
If the components of the vectors are known, $\vec{u} = \langle u_x , u_y , u_z \rangle$ and $\vec{v} = \langle v_x , v_y , v_z \rangle$, the scalar product can be calculated algebraically as:
Theorems on Scalar Products
If $\vec{u}$, $\vec{v}$, and $\vec{w}$ are vectors and $k$ is a scalar, then:
- Commutative: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$
- Distributive: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$
- Magnitude relation: $\vec{u} \cdot \vec{u} = ||\vec{u}||^2$
- Orthogonality: $\vec{u} \cdot \vec{v} = 0$ if and only if $\theta = \pi/2$ (assuming non-zero vectors).
- Scalar multiplication: $(k \vec{u}) \cdot \vec{v} = \vec{u} \cdot (k \vec{v}) = k (\vec{u} \cdot \vec{v})$
- Cauchy-Schwarz Inequality: $|\vec{u} \cdot \vec{v}| \le ||\vec{u}|| \, ||\vec{v}||$
- Triangle Inequality: $|| \vec{u} + \vec{v} || \le ||\vec{u}|| + ||\vec{v}||$
2. Scalar and Vector Projections
In many applications, it is important to find the component of a vector in the direction of another vector. As shown below, vector $\vec{u}$ is projected onto vector $\vec{v}$ by dropping a perpendicular line.
The scalar projection (the component of $\vec{u}$ along $\vec{v}$) is a scalar quantity:
$$ \text{comp}_{\vec{v}}\vec{u} = ||\vec{u}|| \cos \theta $$
The vector projection is a vector quantity obtained by multiplying the scalar projection by the unit vector in the direction of $\vec{v}$:
$$ \text{proj}_{\vec{v}}\vec{u} = ||\vec{u}|| \cos \theta \left(\frac{\vec{v}}{||\vec{v}||}\right) = \left(\frac{\vec{u} \cdot \vec{v}}{||\vec{v}||^2}\right) \vec{v} $$
3. Cross (or Vector) Product of Two Vectors
The cross (or vector) product of two vectors $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$ is a vector quantity defined by the determinant:
$$ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} $$
$$ = \begin{vmatrix} u_y & u_z \\ v_y & v_z \end{vmatrix} \vec{i} - \begin{vmatrix} u_x & u_z \\ v_x & v_z \end{vmatrix} \vec{j} + \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix} \vec{k} $$
The Right-Hand Rule
The cross product $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$. To find its direction, point your index finger in the direction of $\vec{u}$, your middle finger towards $\vec{v}$, and your thumb will point in the direction of $\vec{u} \times \vec{v}$.
Theorems on Cross Products
- Orthogonality: $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$.
- Anti-commutative: $\vec{u} \times \vec{v} = - (\vec{v} \times \vec{u})$
- Parallel condition: $\vec{u} \times \vec{v} = 0$ if and only if $\vec{u}$ and $\vec{v}$ are parallel (assuming non-zero vectors).
- Distributive: $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$
- Scalar multiplication: $(k \vec{u}) \times \vec{v} = \vec{u} \times (k \vec{v}) = k (\vec{u} \times \vec{v})$
- Magnitude relation: $||\vec{u} \times \vec{v}|| = ||\vec{u}|| \, ||\vec{v}|| \sin \theta$
4. Geometric Applications: Area and Volume
Area of a Parallelogram
For a parallelogram defined by adjacent vectors $\vec{AB}$ and $\vec{AD}$, the area is given by the magnitude of their cross product:
$$ \text{Area} = || \vec{AB} \times \vec{AD} || $$
Note: The area of a triangle is exactly half of the corresponding parallelogram's area.
Volume of a Parallelepiped
A parallelepiped is a 3D solid bounded by 6 parallelograms. If it is defined by three vectors sharing a vertex ($\vec{u}, \vec{v}, \vec{w}$), its volume is the absolute value of the scalar triple product:
$$ \text{Volume} = |\vec{u} \cdot (\vec{v} \times \vec{w})| = |\vec{v} \cdot (\vec{w} \times \vec{u})| = |\vec{w} \cdot (\vec{u} \times \vec{v})| $$
Concept Examples
Example 1: Finding the angle between two vectors
Approximate, to the nearest degree, the angle between the vectors $\vec{v} = \langle -2, 3, 1 \rangle$ and $\vec{u} = \langle 0, -1, 4 \rangle$.
Solution:
Express the scalar product of the two vectors using both the algebraic and geometric formulas:
$$ \vec{v} \cdot \vec{u} = (-2)(0) + (3)(-1) + (1)(4) = -3 + 4 = 1 $$
Calculate the magnitudes:
$$ ||\vec{v}|| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14} $$
$$ ||\vec{u}|| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{17} $$
Solve for $\cos\theta$:
$$ \cos\theta = \frac{\vec{v} \cdot \vec{u}}{||\vec{v}|| \, ||\vec{u}||} = \frac{1}{\sqrt{14}\sqrt{17}} $$
$$ \theta = \arccos\left(\frac{1}{\sqrt{14}\sqrt{17}}\right) \approx 86^\circ $$
Example 2: Unknown coordinate of perpendicular vectors
Find $a$ so that the vectors $\langle a, -6, 3 \rangle$ and $\langle 1, 0, -2 \rangle$ are perpendicular.
Solution:
For two vectors to be perpendicular, their scalar (dot) product must equal zero.
$$ \langle a, -6, 3 \rangle \cdot \langle 1, 0, -2 \rangle = a(1) + (-6)(0) + (3)(-2) = 0 $$
$$ a - 6 = 0 $$
$$ a = 6 $$
Example 3: Calculating a cross product
Calculate the cross product of the vectors $\vec{u} = \langle 1, 1, 3 \rangle$ and $\vec{v} = \langle 1, 0, 2 \rangle$.
Solution:
Set up the determinant matrix:
$$ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 3 \\ 1 & 0 & 2 \end{vmatrix} $$
$$ = \begin{vmatrix} 1 & 3 \\ 0 & 2 \end{vmatrix} \vec{i} - \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \vec{k} $$
$$ = ((1)(2) - 0)\vec{i} - ((1)(2) - (1)(3))\vec{j} + ((1)(0) - (1)(1))\vec{k} $$
$$ = 2\vec{i} + 1\vec{j} - 1\vec{k} = \langle 2, 1, -1 \rangle $$
Example 4: Finding perpendicular unit vectors
Find two unit vectors perpendicular to the vectors $\vec{u} = \langle 1, -2, 1 \rangle$ and $\vec{v} = \langle -2, 0, 4 \rangle$.
Solution:
The cross product $\vec{w} = \vec{u} \times \vec{v}$ generates a vector orthogonal to both.
$$ \vec{w} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -2 & 1 \\ -2 & 0 & 4 \end{vmatrix} = (-8 - 0)\vec{i} - (4 - (-2))\vec{j} + (0 - 4)\vec{k} = \langle -8, -6, -4 \rangle $$
Find the magnitude of $\vec{w}$ to normalize it:
$$ ||\vec{w}|| = \sqrt{(-8)^2 + (-6)^2 + (-4)^2} = \sqrt{64 + 36 + 16} = \sqrt{116} = 2\sqrt{29} $$
The first unit vector $\vec{u}_1$ points in the same direction:
$$ \vec{u}_1 = \frac{1}{2\sqrt{29}} \langle -8, -6, -4 \rangle = \left\langle -\frac{4}{\sqrt{29}}, -\frac{3}{\sqrt{29}}, -\frac{2}{\sqrt{29}} \right\rangle $$
The second unit vector $\vec{u}_2$ points in the exact opposite direction:
$$ \vec{u}_2 = -\vec{u}_1 = \left\langle \frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{2}{\sqrt{29}} \right\rangle $$
Example 5: Analyzing vector statements
Explain mathematically why the following statements are not true:
a) $\vec{u} \times \vec{u} = ||\vec{u}||^2$
b) $\vec{u} \cdot (\vec{u} \times \vec{w}) = (\vec{u} \cdot \vec{u}) \times \vec{w}$
Solution:
a) Dimensional Mismatch: The left side ($\vec{u} \times \vec{u}$) is a cross product, meaning the result must be a vector (specifically, the zero vector $\vec{0}$). The right side ($||\vec{u}||^2$) represents a scalar magnitude. A vector and a scalar cannot be equated.
b) Operation Invalidity: On the left side, the cross product creates a vector, which is then dot-producted with $\vec{u}$ resulting in a valid scalar. On the right side, the dot product $(\vec{u} \cdot \vec{u})$ evaluates to a scalar, which is then cross-producted with vector $\vec{w}$. The cross product operation is only defined between two vectors, not a scalar and a vector. Therefore, the right side is mathematically undefined.
Practice Problems & Detailed Solutions
Question 1: Vector Properties
Calculate $\vec{u} \cdot (\vec{u} \times \vec{v})$ given that $\vec{u} = \langle a,b,c \rangle$ and $\vec{v} = \langle d,e,f \rangle$.
Solution:
The cross product $\vec{u} \times \vec{v}$ is a vector that is strictly perpendicular to both original vectors $\vec{u}$ and $\vec{v}$.
Hence, the dot product of two perpendicular vectors ($\vec{u}$ and $\vec{u} \times \vec{v}$) is always equal to $0$.
$$ \vec{u} \cdot (\vec{u} \times \vec{v}) = 0 $$
Question 2: Perpendicular Vectors
Find $k$ so that vectors $\vec{u} = \langle -2,-k,1 \rangle$ and $\vec{v} = \langle 8,-2,-3 \rangle$ are perpendicular.
Solution:
If two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular, then their dot product is equal to $0$.
$$ \vec{u} \cdot \vec{v} = \langle -2, -k, 1 \rangle \cdot \langle 8, -2, -3 \rangle = 0 $$
Expand the dot product to obtain the algebraic equation:
$$ (-2)(8) + (-k)(-2) + (1)(-3) = 0 $$
$$ -16 + 2k - 3 = 0 \implies 2k = 19 $$
$$ k = \frac{19}{2} $$
Question 3: Coplanar Vectors
Find $k$ so that the vectors $\vec{u} = \langle -3,2,-2 \rangle$, $\vec{v} = \langle 2,1,k \rangle$, and $\vec{w} = \langle -1,3,-5 \rangle$ are on the same plane (coplanar).
Solution:
Any two vectors lie on the same plane. If a third vector also lies on this plane, then the volume of the parallelepiped formed by the three vectors is zero. This volume is calculated using the scalar triple product:
$$ \vec{u} \cdot (\vec{v} \times \vec{w}) = 0 $$
This is computed using the determinant of a $3 \times 3$ matrix:
$$ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} -3 & 2 & -2 \\ 2 & 1 & k \\ -1 & 3 & -5 \end{vmatrix} = 0 $$
$$ = -3(-5 - 3k) - 2(-10 + k) - 2(6 + 1) = 21 + 7k $$
For the vectors to be coplanar, set the result to zero:
$$ 21 + 7k = 0 \implies k = -3 $$
Question 4: Angle Between Vectors
Find the angle $\theta$ between the vectors $\vec{u} = \langle 2,0,1 \rangle$ and $\vec{v} = \langle 8,-2,-3 \rangle$.
Solution:
Use the definition of the scalar product: $$ \vec{u} \cdot \vec{v} = \|\vec{u}\|\|\vec{v}\|\cos\theta $$
First, calculate the magnitudes $\|\vec{u}\|$ and $\|\vec{v}\|$:
$$ \|\vec{u}\| = \sqrt{2^2 + 0^2 + 1^2} = \sqrt{5} $$
$$ \|\vec{v}\| = \sqrt{8^2 + (-2)^2 + (-3)^2} = \sqrt{77} $$
Now, calculate the algebraic dot product:
$$ \vec{u} \cdot \vec{v} = (2)(8) + (0)(-2) + (1)(-3) = 13 $$
Isolate $\cos\theta$ and solve:
$$ \cos\theta = \frac{13}{\sqrt{5}\sqrt{77}} = \frac{13}{\sqrt{385}} $$
$$ \theta = \arccos\left(\frac{13}{\sqrt{385}}\right) \approx 48.5^\circ $$
Question 5: Vector Projection
Find the vector projection of $\vec{u} = \langle -1,-1,1 \rangle$ onto $\vec{v} = \langle 2,1,1 \rangle$.
Solution:
The vector projection formula is: $$ \text{proj}_{\vec{v}}\vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\right) \vec{v} $$
Calculate the dot product and the magnitude squared:
$$ \vec{u} \cdot \vec{v} = (-1)(2) + (-1)(1) + (1)(1) = -2 $$
$$ \|\vec{v}\|^2 = 2^2 + 1^2 + 1^2 = 6 $$
Multiply the scalar ratio by the vector $\vec{v}$:
$$ \text{proj}_{\vec{v}}\vec{u} = -\frac{2}{6} \langle 2,1,1 \rangle = \left\langle -\frac{2}{3}, -\frac{1}{3}, -\frac{1}{3} \right\rangle $$
Question 6: Right Triangle from Coordinates
Find the value(s) of $k$ so that the points $A(-1,2,k)$, $B(-3,6,3)$, and $C(1,3,6)$ form the vertices of a right triangle with the $90^\circ$ angle at vertex $A$.
Solution:
For $\triangle ABC$ to be a right triangle at $A$, vectors $\vec{AB}$ and $\vec{AC}$ must be perpendicular. Therefore, their dot product must equal $0$.
$$ \vec{AB} = \langle -3 - (-1), 6 - 2, 3 - k \rangle = \langle -2, 4, 3 - k \rangle $$
$$ \vec{AC} = \langle 1 - (-1), 3 - 2, 6 - k \rangle = \langle 2, 1, 6 - k \rangle $$
Set their dot product to zero:
$$ \langle -2, 4, 3 - k \rangle \cdot \langle 2, 1, 6 - k \rangle = 0 $$
$$ -4 + 4 + (3 - k)(6 - k) = 0 $$
$$ (3 - k)(6 - k) = 0 $$
This yields two valid solutions: $k = 3$ and $k = 6$.
Question 7: Solving for an Unknown Vector
Given vector $\vec{v} = \langle 3,-1,-2 \rangle$, find a vector $\vec{u}$ such that $\vec{v} \times \vec{u} = \langle 4,2,5 \rangle$ and $\|\vec{u}\| = 3$.
Solution:
Let $a, b, c$ be the components of vector $\vec{u}$. Evaluate the cross product:
$$ \vec{v} \times \vec{u} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & -1 & -2 \\ a & b & c \end{vmatrix} = (-c + 2b)\vec{i} - (3c + 2a)\vec{j} + (3b + a)\vec{k} $$
Set the components equal to $\langle 4, 2, 5 \rangle$:
1) $-c + 2b = 4$
2) $-2a - 3c = 2$
3) $3b + a = 5$
These equations are dependent. Let $a = t$. From eq 2: $c = \frac{-2 - 2t}{3}$. From eq 3: $b = \frac{5 - t}{3}$.
Apply the magnitude constraint $\|\vec{u}\| = 3 \implies a^2 + b^2 + c^2 = 9$:
$$ t^2 + \left(\frac{5 - t}{3}\right)^2 + \left(\frac{-2 - 2t}{3}\right)^2 = 9 $$
Multiply by 9 and expand:
$$ 9t^2 + (25 - 10t + t^2) + (4 + 8t + 4t^2) = 81 $$
$$ 14t^2 - 2t - 52 = 0 \implies 7t^2 - t - 26 = 0 $$
Factoring yields $t = 2$ and $t = -\frac{13}{7}$. Substituting these back gives two possible vectors:
If $t = 2$: $\vec{u}_1 = \langle 2, 1, -2 \rangle$
If $t = -\frac{13}{7}$: $\vec{u}_2 = \langle -\frac{13}{7}, \frac{16}{7}, \frac{4}{7} \rangle$
Question 8: Parallelogram Area & Vertices
Points $A(4,6,2)$, $B(2,2,4)$, $C(-2,-3,1)$ and an unknown point $D$ form a parallelogram.
a) Find the coordinates of point $D$.
b) Find the area of the parallelogram.
Solution:
a) Let $D = (a, b, c)$. For $ABCD$ to be a parallelogram, vector $\vec{AB}$ must equal vector $\vec{DC}$.
$$ \vec{AB} = \langle 2 - 4, 2 - 6, 4 - 2 \rangle = \langle -2, -4, 2 \rangle $$
$$ \vec{DC} = \langle -2 - a, -3 - b, 1 - c \rangle $$
Equating components:
$$ -2 - a = -2 \implies a = 0 $$
$$ -3 - b = -4 \implies b = 1 $$
$$ 1 - c = 2 \implies c = -1 $$
Therefore, the coordinates of point $D$ are $(0, 1, -1)$.
b) The area of the parallelogram is given by the magnitude of the cross product of two adjacent sides: $\text{Area} = \|\vec{AB} \times \vec{BC}\|$.
First, find $\vec{BC}$:
$$ \vec{BC} = \langle -2 - 2, -3 - 2, 1 - 4 \rangle = \langle -4, -5, -3 \rangle $$
Next, calculate the cross product $\vec{AB} \times \vec{BC}$:
$$ \vec{AB} \times \vec{BC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & -4 & 2 \\ -4 & -5 & -3 \end{vmatrix} $$
$$ = ((-4)(-3) - (2)(-5))\vec{i} - ((-2)(-3) - (2)(-4))\vec{j} + ((-2)(-5) - (-4)(-4))\vec{k} $$
$$ = (12 + 10)\vec{i} - (6 + 8)\vec{j} + (10 - 16)\vec{k} = 22\vec{i} - 14\vec{j} - 6\vec{k} $$
Finally, compute the magnitude to find the area:
$$ \text{Area} = \sqrt{22^2 + (-14)^2 + (-6)^2} = \sqrt{484 + 196 + 36} = \sqrt{716} = 2\sqrt{179} $$
Question 9: Angle in a Cube
In a cube with side length 2, find the angle between the primary body diagonals $AG$ and $BH$.
Solution:
Assume the cube is aligned with the axes. The components of the diagonal vectors are:
$$ \vec{AG} = \langle 2, 2, 2 \rangle \quad \text{and} \quad \vec{BH} = \langle -2, 2, 2 \rangle $$
Using the dot product formula:
$$ \cos\theta = \frac{\vec{AG} \cdot \vec{BH}}{\|\vec{AG}\|\|\vec{BH}\|} = \frac{(2)(-2) + (2)(2) + (2)(2)}{\sqrt{12}\sqrt{12}} = \frac{4}{12} = \frac{1}{3} $$
$$ \theta = \arccos\left(\frac{1}{3}\right) \approx 70.5^\circ $$
Question 10: Orthogonal to a Plane
Find a vector that is orthogonal to the plane containing the points $A(1,2,-3)$, $B(0,-2,1)$, and $C(-2,0,1)$.
Solution:
First, compute two vectors lying within the plane:
$$ \vec{AB} = \langle 0-1, -2-2, 1-(-3) \rangle = \langle -1, -4, 4 \rangle $$
$$ \vec{AC} = \langle -2-1, 0-2, 1-(-3) \rangle = \langle -3, -2, 4 \rangle $$
The cross product of these two vectors generates a vector strictly normal (orthogonal) to the plane:
$$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & -4 & 4 \\ -3 & -2 & 4 \end{vmatrix} $$
$$ = ((-4)(4) - (4)(-2))\vec{i} - ((-1)(4) - (4)(-3))\vec{j} + ((-1)(-2) - (-4)(-3))\vec{k} $$
$$ = (-16 + 8)\vec{i} - (-4 + 12)\vec{j} + (2 - 12)\vec{k} = \langle -8, -8, -10 \rangle $$
Question 11: Area of a 3D Triangle
Find the area of the triangle whose vertices are the points $A(1, 0, -3)$, $B(1, -2, 0)$, and $C(0, 2, 1)$.
Solution:
The area $A$ of a triangle formed by two vectors is half the magnitude of their cross product. Let's use vectors $\vec{AB}$ and $\vec{AC}$.
$$ \vec{AB} = \langle 0, -2, 3 \rangle \quad \text{and} \quad \vec{AC} = \langle -1, 2, 4 \rangle $$
$$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & -2 & 3 \\ -1 & 2 & 4 \end{vmatrix} = (-8 - 6)\vec{i} - (0 - (-3))\vec{j} + (0 - 2)\vec{k} = \langle -14, -3, -2 \rangle $$
$$ Area = \frac{1}{2}\|\vec{AB} \times \vec{AC}\| = \frac{1}{2}\sqrt{(-14)^2 + (-3)^2 + (-2)^2} = \frac{1}{2}\sqrt{196 + 9 + 4} $$
$$ Area = \frac{1}{2}\sqrt{209} \text{ units}^2 $$
Question 12: Volume of a Parallelepiped
Find the volume of the parallelepiped defined by the origin vectors $\vec{u} = \langle -3, 0, 7 \rangle$, $\vec{v} = \langle -8, 0, 0 \rangle$, and $\vec{w} = \langle 0, -9, 0 \rangle$.
Solution:
The volume $V$ is the absolute value of the scalar triple product: $V = |\vec{u} \cdot (\vec{v} \times \vec{w})|$.
$$ \vec{v} \times \vec{w} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -8 & 0 & 0 \\ 0 & -9 & 0 \end{vmatrix} = (0)\vec{i} - (0)\vec{j} + (72)\vec{k} = \langle 0, 0, 72 \rangle $$
Dot product with $\vec{u}$:
$$ \vec{u} \cdot (\vec{v} \times \vec{w}) = \langle -3, 0, 7 \rangle \cdot \langle 0, 0, 72 \rangle = 0 + 0 + (7)(72) = 504 $$
$$ V = |504| = 504 \text{ units}^3 $$