Vectors & 3‑D Geometry
Engineering Services – Mathematics
(For quick reference & revision)
1. Simple Explanation
- Vector – an object that has both magnitude and direction.
- In 3‑D space a vector is usually written
[ \mathbf{v}= \langle v_x,,v_y,,v_z\rangle ] or using unit‑vector notation
[ \mathbf{v}=v_x,\mathbf{i}+v_y,\mathbf{j}+v_z,\mathbf{k} ] - Vectors can represent displacements, forces, velocity, etc.
- Operations on vectors (addition, subtraction, scalar multiplication) obey the usual algebraic rules but are geometrically meaningful.
2. Key Points
| Topic | What to Know | Quick Formulas |
|---|---|---|
| Magnitude | Length of a vector | (|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2}) |
| Unit Vector | Vector of length 1 in same direction | (\hat{\mathbf{v}}=\dfrac{\mathbf{v}}{|\mathbf{v}|}) |
| Addition/Subtraction | Component‑wise | (\mathbf{u}+\mathbf{v}=\langle u_x+v_x,;u_y+v_y,;u_z+v_z\rangle) |
| Scalar Multiplication | Scale every component | (k\mathbf{v}=\langle kv_x, kv_y, kv_z\rangle) |
| Dot Product | Measures parallelism | (\mathbf{u}\cdot\mathbf{v}=u_xv_x+u_yv_y+u_zv_z) |
| Alternative: (\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta) | ||
| Cross Product | Gives a vector orthogonal to both | (\mathbf{u}\times\mathbf{v}=\langle u_yv_z-u_zv_y,;u_zv_x-u_xv_z,;u_xv_y-u_yv_x\rangle) |
| Magnitude = area of parallelogram | (|\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta) | |
| Line in 3‑D | Parametric form | (\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{d}) |
| Plane | Cartesian form | (ax+by+cz=d) |
| Normal vector (\mathbf{n}=\langle a,b,c\rangle) | ||
| Distance | Point to plane | (\displaystyle \text{dist}=\frac{ |
| Point to line | (\displaystyle \frac{|\mathbf{P}-\mathbf{A}\times\mathbf{d}|}{|\mathbf{d}|}) | |
| Intersection | Line‑plane: substitute parametric line into plane equation. | |
| Line‑line: solve for parameters, check if they’re equal (intersect) or not (skew). |
3. Worked Example
Problem
Find the distance from point (P(2,,5,,-1)) to the plane
(3x-4y+2z=7).
Solution Steps
-
Identify the plane’s normal
(\mathbf{n}=\langle 3,,-4,,2\rangle) -
Plug into distance formula
[ d=\frac{|3(2)+(-4)(5)+2(-1)-7|}{\sqrt{3^2+(-4)^2+2^2}} ] -
Compute numerator
[ |6-20-2-7| = |-23| = 23 ] -
Compute denominator
[ \sqrt{9+16+4}= \sqrt{29} ] -
Result
[ d=\frac{23}{\sqrt{29}}\approx 4.28 ]
So, the shortest distance from (P) to the plane is ≈ 4.28 units.
4. Quick Revision Summary
- Vectors: (\mathbf{v}=\langle v_x,v_y,v_z\rangle).
- Magnitude: (|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2}).
- Unit vector: (\hat{\mathbf{v}}=\mathbf{v}/|\mathbf{v}|).
- Dot product: (\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta).
- Cross product: (\mathbf{u}\times\mathbf{v}=\langle u_yv_z-u_zv_y,\dots\rangle).
- Line equation: (\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{d}).
- Plane equation: (