Vectors and 3‑D Geometry – definitions, operations, lines, planes & applications

1. Vectors in Three‑Dimensional Space

A vector $\mathbf{v}$ in ${\mathbb{R}}^3$ is an ordered triple $(v_x,v_y,v_z)$. It is visualised as an arrow from the origin $O(0,0,0)$ to the point $P(v_x,v_y,v_z)$.

Magnitude (length)
$$|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2}.$$

Direction cosines are the cosines of the angles that $\mathbf{v}$ makes with the positive $x$, $y$, $z$ axes: $$\cos \alpha =\frac{v_x}{|\mathbf{v}|},\cos \beta =\frac{v_y}{|\mathbf{v}|},\cos \gamma =\frac{v_z}{|\mathbf{v}|}.$$

Unit vector in the direction of $\mathbf{v}$: $$\hat{\mathbf{v}}=\frac{\mathbf{v}}{|\mathbf{v}|}.$$

2. Vector Operations

2.1 Dot product and angle

$$\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos \theta ⟹\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$$

2.2 Cross product and perpendicular vector

$$\mathbf{a}\times \mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin \theta \hat{\mathbf{n}},$$ where $\hat{\mathbf{n}}$ is a unit vector normal to the plane containing $\mathbf{a},\mathbf{b}$ (right‑hand rule).

2.3 Mixed product – volume of a parallelepiped

The absolute value $|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$ equals the volume $V$ of the parallelepiped built on $\mathbf{a},\mathbf{b},\mathbf{c}$.

3. Lines and Planes in Space

3.1 Vector equation of a line

Through point $P_0(x_0,y_0,z_0)$ with direction vector $\mathbf{d}$: $$\mathbf{r}={\mathbf{r}}_0+t\mathbf{d},t\in \mathbb{R},$$ or component form $$\frac{x-x_0}{d_x}=\frac{y-y_0}{d_y}=\frac{z-z_0}{d_z}=t.$$

3.2 Plane equation

A plane with normal vector $\mathbf{n}=(a,b,c)$ passing through point $P_0(x_0,y_0,z_0)$: $$\mathbf{n}\cdot (\mathbf{r}-{\mathbf{r}}_0)=0⟹a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$

If the normal is known, the scalar form is $$ax+by+cz+d=0,$$ where $d=-(a{x}_0+b{y}_0+c{z}_0)$.

3.3 Distance formulas

• Point to plane $ax+by+cz+d=0$: $$\text{dist}=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}.$$

• Point to line (using vector projection): $$\text{dist}=\frac{|\mathbf{AP}\times \mathbf{d}|}{|\mathbf{d}|},$$ where $\mathbf{A}$ is a point on the line, $\mathbf{P}$ the external point, and $\mathbf{d}$ the direction vector.

4. Worked Examples

Example 1 – Unit vector perpendicular to a plane

Problem: Find a unit vector normal to the plane through
$P(1,-1,0),Q(2,1,-1),R(-1,1,2)$.

Solution:

1. Form two direction vectors in the plane:
   $$\begin{gathered} \mathbf{u}=\overrightarrow{PQ}=(2-1,1-(-1),-1-0)=(1,2,-1), \\ \mathbf{v}=\overrightarrow{PR}=(-1-1,1-(-1),2-0)=(-2,2,2). \end{gathered}$$

2. Compute the cross product $\mathbf{n}=\mathbf{u}\times \mathbf{v}$: $$\mathbf{n}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& -1\\ -2& 2& 2\end{array}\right|=(2\cdot 2-(-1)\cdot 2)\mathbf{i}-(1\cdot 2-(-1)(-2))\mathbf{j}+(1\cdot 2-2(-2))\mathbf{k}=(4+2)\mathbf{i}-(2-2)\mathbf{j}+(2+4)\mathbf{k}=(6,0,6).$$

3. Simplify: $\mathbf{n}=(6,0,6)=6(1,0,1)$.

4. Magnitude $|\mathbf{n}|=6\sqrt{1^2+0^2+1^2}=6\sqrt{2}$.

5. Unit normal vector: $$\hat{\mathbf{n}}=\frac{\mathbf{n}}{|\mathbf{n}|}=\frac{1}{\sqrt{2}}(1,0,1).$$

Result: $\hat{\mathbf{n}}=\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$.

Example 2 – Volume of a parallelepiped

Problem: Find the volume of the parallelepiped whose concurrent edges are represented by
$\mathbf{a}=\mathbf{i}-2\mathbf{j}+3\mathbf{k},\mathbf{b}=2\mathbf{i}+\mathbf{j}-\mathbf{k},\mathbf{c}=-\mathbf{i}+\mathbf{j}+2\mathbf{k}$.

Solution:

Compute the scalar triple product: $$V=|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|=\left|\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& -1\\ -1& 1& 2\end{array}\right|\right|.$$

Evaluating the determinant: $$\begin{array}{rl}\Delta & =1\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|-(-2)\left|\begin{array}{cc}2& -1\\ -1& 2\end{array}\right|+3\left|\begin{array}{cc}2& 1\\ -1& 1\end{array}\right|\\ & =1(1\cdot 2-(-1)\cdot 1)+2(2\cdot 2-(-1)(-1))+3(2\cdot 1-1(-1))\\ & =1(2+1)+2(4-1)+3(2+1)\\ & =3+2\cdot 3+3\cdot 3=3+6+9=18.\end{array}$$

Thus $V=|18|=18$ cubic units.

Example 3 – Angle between two vectors

Problem: Find the angle between $\mathbf{u}=4\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\mathbf{v}=\mathbf{i}+\mathbf{j}-\mathbf{k}$.

Solution:

$$\mathbf{u}\cdot \mathbf{v}=4\cdot 1+(-2)\cdot 1+1\cdot (-1)=4-2-1=1.$$

$$|\mathbf{u}|=\sqrt{4^2+(-2{)}^2+1^2}=\sqrt{16+4+1}=\sqrt{21},|\mathbf{v}|=\sqrt{1^2+1^2+(-1{)}^2}=\sqrt{3}.$$

$$\cos \theta =\frac{1}{\sqrt{21}\sqrt{3}}=\frac{1}{\sqrt{63}}.$$

$$\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{63}}\right)\approx 81.8^{\circ }.$$

5. Comparison of Dot and Cross Products

flowchart LR
    D["Dot product"] -->|"Result"| S["Scalar (gives magnitude of projection)"]
    C["Cross product"] -->|"Result"| V["Vector (perpendicular, gives area)"]
    D -->|"Uses"| A["Angle between vectors"]
    C -->|"Uses"| B["Normal to a plane"]

• Dot product is commutative ($\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$), useful for angles and work.
• Cross product is anti‑commutative ($\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$), gives a vector orthogonal to the original pair, used for torque, area, and normals.

6. Applications in the Real World

6.1 GPS Navigation (Google Maps, NTC)

The position of a GPS receiver is obtained by solving three‑dimensional vector equations derived from satellite–receiver distance vectors. Each satellite provides a line‑of‑sight vector; the intersection of three such lines yields the receiver’s coordinates.

6.2 Drone Delivery (Pathao, Daraz)

A delivery drone follows a flight path defined by a parametric vector line $\mathbf{r}(t)={\mathbf{r}}_0+t\mathbf{d}$. Obstacles are modelled as planes; the drone’s autopilot computes the shortest distance from its current point to each obstacle plane using the point‑to‑plane formula, then adjusts its direction vector to avoid collisions.

6.3 3‑D Modelling in E‑Commerce (Daraz product visualiser)

When a product is displayed in 3‑D, the rotation of the model is performed by applying rotation matrices to the position vectors of its vertices. The normal vectors of each face, obtained via cross products, determine lighting and shading, giving a realistic view.

Worked real‑world tie‑in:
A Daraz order queue can be visualised as a vector of pending orders $\mathbf{q}=(q_1,q_2,q_3)$ where each component represents orders in a regional warehouse. The dot product of $\mathbf{q}$ with a priority vector $\mathbf{p}=(3,2,1)$ yields a scalar “urgency score” that the system uses to allocate delivery resources.

7. Advantages & Limitations

8. Summary of Key Formulas

9. Exam tip

TU exams frequently ask for “vector method” proofs. Memorise the determinant form of the scalar triple product and the right‑hand rule for cross products; they let you write the answer in a single line. When a question gives three points, always compute two side vectors first, then use the cross product to obtain the normal—this avoids algebraic errors in plane equations. For distance problems, plug the normal vector directly into the point‑to‑plane formula; it’s faster than projecting onto the line.

GPS satellites provide 3‑D position vectors for navigation (Image: U.S. Govt Source, Public domain, via Wikimedia Commons)