Applications of Vectors in M2
Hey everyone! Welcome to the study notes for "Applications of Vectors". You've learned what vectors are and how to do basic operations like adding and multiplying them. Now, let's unlock their true power! In this chapter, we'll see how vectors are not just abstract arrows but powerful tools for solving real geometric problems in 2D and 3D space. We'll learn how to find angles, project shadows, calculate areas, and much more.
Don't worry if this sounds complicated. We'll break everything down into simple, step-by-step pieces. Think of it like adding new tools to your maths toolkit. Let's get started!
1. Division of a Line Segment
Imagine you have a line segment from point A to point B. What if you want to find the exact coordinates of a point P that is, say, halfway along the line? Or maybe two-thirds of the way? This is where the section formula comes in, and it's super useful!
Internal Division
This is the most common case. We want to find a point P that lies between A and B and divides the line segment AB in a specific ratio, let's say m : n.
Let $$\vec{a}$$ and $$\vec{b}$$ be the position vectors of points A and B. The position vector of point P, which divides AB in the ratio m : n, is given by:
Section Formula (Internal Division):
$$ \vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n} $$Memory Aid: The "Cross-Multiply" Trick
To remember the formula, think of it like this: the ratio part 'n' multiplies with the vector 'a' (the one further away), and the ratio part 'm' multiplies with the vector 'b' (the other one further away). Then, you divide by the sum of the ratios (m+n).
Example: Let A = (1, 2, 3) and B = (5, 6, 7). Find the point P that divides AB internally in the ratio 1 : 3.Step 1: Identify your vectors and ratio.
$$\vec{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$, $$\vec{b} = \begin{pmatrix} 5 \\ 6 \\ 7 \end{pmatrix}$$
m = 1, n = 3
Step 2: Apply the formula.
$$ \vec{p} = \frac{3\vec{a} + 1\vec{b}}{1+3} = \frac{3\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 1\begin{pmatrix} 5 \\ 6 \\ 7 \end{pmatrix}}{4} $$
Step 3: Calculate the result.
$$ \vec{p} = \frac{\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} + \begin{pmatrix} 5 \\ 6 \\ 7 \end{pmatrix}}{4} = \frac{\begin{pmatrix} 8 \\ 12 \\ 16 \end{pmatrix}}{4} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} $$
So, the coordinates of point P are (2, 3, 4).
A Special Case: The Midpoint
The midpoint is just a point that divides the line segment in a 1 : 1 ratio. If we put m=1 and n=1 into the formula:
$$ \vec{p} = \frac{1\vec{a} + 1\vec{b}}{1+1} = \frac{\vec{a} + \vec{b}}{2} $$This is simply the average of the two vectors!
Key Takeaway
To find a point P dividing line segment AB in the ratio m:n, use the formula $$ \vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n} $$. Remember the "cross-multiply" trick to get the numerator right!
2. Parallelism and Orthogonality (Perpendicularity)
Vectors make it very easy to check the relationship between lines or directions. Are they parallel? Are they at a right angle? Let's see how.
Parallel Vectors
Two vectors are parallel if they point in the same direction or in exact opposite directions. Think of two parallel streets on a map.
The Rule: Two non-zero vectors $$\vec{a}$$ and $$\vec{b}$$ are parallel if one is a scalar multiple of the other.
$$ \vec{a} = k \vec{b} $$Here, 'k' is just a real number (a scalar).
- If k > 0, they are in the same direction.
- If k < 0, they are in the opposite direction.
Let's see if we can find a 'k' such that $$\vec{a} = k\vec{b}$$.
Comparing the i components: $$2 = k(-3) \implies k = -2/3$$
Comparing the j components: $$4 = k(-6) \implies k = -2/3$$
Comparing the k components: $$-6 = k(9) \implies k = -2/3$$
Since we get the same value of k for all components, the vectors are parallel! Because k is negative, they point in opposite directions.
Orthogonal (Perpendicular) Vectors
Two vectors are orthogonal (a fancy word for perpendicular) if the angle between them is 90°.
The Rule: Two non-zero vectors $$\vec{a}$$ and $$\vec{b}$$ are orthogonal if and only if their dot product is zero.
$$ \vec{a} \cdot \vec{b} = 0 $$Why does this work?
Remember the dot product formula: $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$$. If the angle $$\theta$$ is 90°, then $$\cos(90°) = 0$$. This makes the entire dot product zero!
Example: Show that the vectors $$\vec{u} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$$ and $$\vec{v} = 4\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}$$ are orthogonal.We just need to calculate their dot product.
$$ \vec{u} \cdot \vec{v} = (3)(4) + (-2)(5) + (1)(-2) $$
$$ = 12 - 10 - 2 $$
$$ = 0 $$
Since the dot product is 0, the vectors are orthogonal.
Key Takeaway
Parallel Check: Is $$\vec{a} = k\vec{b}$$? (Are the components proportional?)
Perpendicular Check: Is $$\vec{a} \cdot \vec{b} = 0$$? (Is the dot product zero?)
3. Finding the Angle Between Two Vectors
We just used the dot product to check for 90° angles, but we can use it to find ANY angle between two vectors. This is one of the most powerful applications of the dot product.
By rearranging the dot product formula, we get:
$$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} $$Step-by-Step Guide to Finding the Angle
Step 1: Calculate the dot product ($$\vec{a} \cdot \vec{b}$$).
Step 2: Calculate the magnitude of each vector ($$|\vec{a}|$$ and $$|\vec{b}|$$).
Step 3: Plug these three values into the formula to find $$\cos\theta$$.
Step 4: Use the inverse cosine function ($$\theta = \arccos(\dots)$$) on your calculator to find the angle $$\theta$$.
Step 1: Dot Product
$$ \vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4 $$
Step 2: Magnitudes
$$ |\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3 $$
$$ |\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36+9+4} = \sqrt{49} = 7 $$
Step 3: Formula
$$ \cos\theta = \frac{4}{(3)(7)} = \frac{4}{21} $$
Step 4: Find Angle
$$ \theta = \arccos\left(\frac{4}{21}\right) \approx 79.0^\circ $$
The angle between the two vectors is about 79.0 degrees.
Quick Review Box
Angle Formula: $$ \theta = \arccos\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right) $$
4. Projection of a Vector
What is a projection? Imagine the sun is directly overhead. Your shadow on the ground is your "projection" onto the ground. In vectors, we do something similar. The projection of vector $$\vec{a}$$ onto vector $$\vec{b}$$ is like finding the "shadow" that $$\vec{a}$$ casts along the direction of $$\vec{b}$$ .
There are two types of projection: scalar and vector.
Scalar Projection (The Length of the Shadow)
This tells you the magnitude or length of the shadow. It's just a number.
The formula for the scalar projection of $$\vec{a}$$ onto $$\vec{b}$$ is:
$$ \text{comp}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $$Vector Projection (The Shadow as a Vector)
This gives you the actual shadow vector. It has both magnitude and direction.
To get the vector, we take the length (scalar projection) and multiply it by the unit vector in the direction of $$\vec{b}$$ (which is $$\frac{\vec{b}}{|\vec{b}|}$$).
The formula for the vector projection of $$\vec{a}$$ onto $$\vec{b}$$ is:
$$ \text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right) \frac{\vec{b}}{|\vec{b}|} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} $$Common Mistake Alert!
When projecting $$\vec{a}$$ onto $$\vec{b}$$, the vector in the denominator is always $$\vec{b}$$. The vector you're projecting *onto* is the important one for the direction and the denominator.
Example: Find the scalar and vector projection of $$\vec{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$$ onto $$\vec{b} = 4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}$$.Step 1: Find the dot product and magnitude of $$\vec{b}$$.
$$ \vec{a} \cdot \vec{b} = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32 $$
$$ |\vec{b}| = \sqrt{4^2 + 5^2 + 6^2} = \sqrt{16 + 25 + 36} = \sqrt{77} $$
Step 2: Calculate Scalar Projection.
$$ \text{Scalar Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{32}{\sqrt{77}} $$
Step 3: Calculate Vector Projection.
We also need $$|\vec{b}|^2 = 77$$.
$$ \text{Vector Projection} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \frac{32}{77}(4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) $$
$$ = \frac{128}{77}\mathbf{i} + \frac{160}{77}\mathbf{j} + \frac{192}{77}\mathbf{k} $$
Key Takeaway
Projection is like finding a vector's shadow.
- Scalar Projection (Length): $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$
- Vector Projection (Vector): $$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$$
5. Area of a Triangle
Here comes another cool trick! We can use the cross product to find the area of a triangle in 3D space instantly.
Did you know?
The magnitude of the cross product of two vectors, $$|\vec{a} \times \vec{b}|$$, is equal to the area of the parallelogram formed by those two vectors. Since a triangle is just half of a parallelogram, its area is half the magnitude of the cross product!
The Rule: If a triangle is formed by two vectors $$\vec{a}$$ and $$\vec{b}$$ starting from the same point, its area is:
$$ \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| $$Step-by-Step Guide to Finding Triangle Area
Step 1: Find two vectors that form the triangle. IMPORTANT: They must start from the same vertex. For a triangle with vertices P, Q, and R, you can use vectors $$\vec{PQ}$$ and $$\vec{PR}$$.
Step 2: Calculate their cross product ($$\vec{a} \times \vec{b}$$).
Step 3: Find the magnitude of the resulting vector ($$|\vec{a} \times \vec{b}|$$).
Step 4: Divide the magnitude by 2.
Step 1: Find two vectors from the same point (let's use P).
$$ \vec{PQ} = \vec{q} - \vec{p} = (2-1)\mathbf{i} + (3-1)\mathbf{j} + (4-1)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} $$
$$ \vec{PR} = \vec{r} - \vec{p} = (4-1)\mathbf{i} + (3-1)\mathbf{j} + (2-1)\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} $$
Step 2: Calculate the cross product, $$\vec{PQ} \times \vec{PR}$$.
$$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} $$
$$ = \mathbf{i}(2 \cdot 1 - 3 \cdot 2) - \mathbf{j}(1 \cdot 1 - 3 \cdot 3) + \mathbf{k}(1 \cdot 2 - 2 \cdot 3) $$
$$ = \mathbf{i}(2 - 6) - \mathbf{j}(1 - 9) + \mathbf{k}(2 - 6) $$
$$ = -4\mathbf{i} + 8\mathbf{j} - 4\mathbf{k} $$
Step 3: Find the magnitude of this new vector.
$$ |-4\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}| = \sqrt{(-4)^2 + 8^2 + (-4)^2} = \sqrt{16 + 64 + 16} = \sqrt{96} $$
(Hint: $$\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$$)
Step 4: Divide by 2.
$$ \text{Area} = \frac{1}{2}\sqrt{96} = \frac{4\sqrt{6}}{2} = 2\sqrt{6} $$
The area of the triangle is $$2\sqrt{6}$$ square units.
Key Takeaway
To find the area of a triangle with vertices P, Q, R, find two vectors like $$\vec{PQ}$$ and $$\vec{PR}$$, and then use the formula $$ \text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}| $$.