How to Solve: Vectors in 3D (Magnitude, Scalar Product, Angle Between Vectors)

How to Solve: Vectors in 3D (Magnitude, Scalar Product, Angle Between Vectors)

Introduction

"Mastering 3D vectors unlocks 10–15% of your A-Level Mechanics paper—and real-world problems like drone navigation, robotics, and even video game physics. One angle calculation can be worth 6 marks, so let’s make sure you get it right every time."

What You Need To Know First

1. 2D vectors: You must know how to find magnitude and direction in 2D (e.g., ( \mathbf{v} = (3, 4) ) has magnitude 5).
2. Pythagoras’ theorem: Used to calculate distances in 3D.
3. Trigonometry (cosine rule): Needed to find angles between vectors.

If you’re shaky on these, pause and review them first.

Key Vocabulary

Formulas To Know

1. Magnitude of a 3D Vector
   [
   |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}
   ]
2. ( x, y, z ): Components of the vector ( \mathbf{v} = (x, y, z) ).

3. MEMORISE THIS – It’s the 3D version of Pythagoras.

4. Scalar (Dot) Product
   [
   \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z
   ]

5. ( \mathbf{a} = (a_x, a_y, a_z) ), ( \mathbf{b} = (b_x, b_y, b_z) ).

6. MEMORISE THIS – It’s the sum of the products of corresponding components.

7. Angle Between Two Vectors
   [
   \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}
   ]

8. ( \theta ): Angle between ( \mathbf{a} ) and ( \mathbf{b} ).

9. MEMORISE THIS – It links the scalar product to the angle.

10. Alternative Scalar Product Formula (Using Angle)
    [
    \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta
    ]

11. Given on exam sheet (but you must know how to use it).

Step-by-Step Method

1. Finding the Magnitude of a 3D Vector

Steps: 1. Write down the vector in component form: ( \mathbf{v} = (x, y, z) ). 2. Square each component: ( x^2, y^2, z^2 ). 3. Add the squares: ( x^2 + y^2 + z^2 ). 4. Take the square root of the sum: ( \sqrt{x^2 + y^2 + z^2} ).

Example: Find ( |\mathbf{v}| ) where ( \mathbf{v} = (1, -2, 3) ). 1. ( \mathbf{v} = (1, -2, 3) ). 2. ( 1^2 = 1 ), ( (-2)^2 = 4 ), ( 3^2 = 9 ). 3. ( 1 + 4 + 9 = 14 ). 4. ( |\mathbf{v}| = \sqrt{14} ).

2. Calculating the Scalar Product

Steps: 1. Write both vectors in component form: ( \mathbf{a} = (a_x, a_y, a_z) ), ( \mathbf{b} = (b_x, b_y, b_z) ). 2. Multiply corresponding components: ( a_x b_x ), ( a_y b_y ), ( a_z b_z ). 3. Add the results: ( a_x b_x + a_y b_y + a_z b_z ).

Example: Find ( \mathbf{a} \cdot \mathbf{b} ) where ( \mathbf{a} = (2, -1, 3) ) and ( \mathbf{b} = (4, 5, 0) ). 1. ( \mathbf{a} = (2, -1, 3) ), ( \mathbf{b} = (4, 5, 0) ). 2. ( 2 \times 4 = 8 ), ( -1 \times 5 = -5 ), ( 3 \times 0 = 0 ). 3. ( 8 + (-5) + 0 = 3 ).

3. Finding the Angle Between Two Vectors

Steps: 1. Find the scalar product ( \mathbf{a} \cdot \mathbf{b} ). 2. Find the magnitudes ( |\mathbf{a}| ) and ( |\mathbf{b}| ). 3. Plug into the formula: ( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} ). 4. Take the inverse cosine (( \cos^{-1} )) of both sides to find ( \theta ).

Example: Find the angle between ( \mathbf{a} = (1, 0, -1) ) and ( \mathbf{b} = (0, 1, 1) ). 1. ( \mathbf{a} \cdot \mathbf{b} = (1 \times 0) + (0 \times 1) + (-1 \times 1) = -1 ). 2. ( |\mathbf{a}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} ).
( |\mathbf{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} ). 3. ( \cos \theta = \frac{-1}{\sqrt{2} \times \sqrt{2}} = \frac{-1}{2} ). 4. ( \theta = \cos^{-1}\left(-\frac{1}{2}\right) = 120° ).

Worked Examples

Example 1 – Basic: Magnitude and Scalar Product

Question: Given ( \mathbf{p} = (3, -2, 6) ) and ( \mathbf{q} = (1, 4, -2) ), find: a) ( |\mathbf{p}| ) b) ( \mathbf{p} \cdot \mathbf{q} )

Solution: a) ( |\mathbf{p}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 ). b) ( \mathbf{p} \cdot \mathbf{q} = (3 \times 1) + (-2 \times 4) + (6 \times -2) = 3 - 8 - 12 = -17 ).

What we did and why: - For magnitude, we used the 3D Pythagoras formula. - For the scalar product, we multiplied corresponding components and added them. This gives a single number representing how much one vector "points" in the direction of the other.

Example 2 – Medium: Angle Between Vectors

Question: Find the angle between ( \mathbf{u} = (2, 1, -3) ) and ( \mathbf{v} = (-1, 4, 2) ). Give your answer to 1 decimal place.

Solution: 1. ( \mathbf{u} \cdot \mathbf{v} = (2 \times -1) + (1 \times 4) + (-3 \times 2) = -2 + 4 - 6 = -4 ). 2. ( |\mathbf{u}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} ).
( |\mathbf{v}| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} ). 3. ( \cos \theta = \frac{-4}{\sqrt{14} \times \sqrt{21}} = \frac{-4}{\sqrt{294}} ).
Simplify ( \sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6} ).
So, ( \cos \theta = \frac{-4}{7\sqrt{6}} ). 4. ( \theta = \cos^{-1}\left(\frac{-4}{7\sqrt{6}}\right) \approx 106.6° ).

What we did and why: - We used the scalar product and magnitudes to find ( \cos \theta ). - Simplifying the denominator made the calculation cleaner. - The negative scalar product tells us the angle is obtuse (> 90°).

Example 3 – Exam-Style: Disguised Problem

Question: A drone moves from point ( A(1, 2, 3) ) to point ( B(4, 6, 8) ). A wind force acts in the direction of vector ( \mathbf{w} = (1, 1, 1) ). a) Find the displacement vector ( \overrightarrow{AB} ). b) Calculate the angle between the drone’s path and the wind force. Give your answer in degrees to 1 decimal place.

Solution: a) ( \overrightarrow{AB} = B - A = (4 - 1, 6 - 2, 8 - 3) = (3, 4, 5) ). b) 1. ( \overrightarrow{AB} \cdot \mathbf{w} = (3 \times 1) + (4 \times 1) + (5 \times 1) = 3 + 4 + 5 = 12 ). 2. ( |\overrightarrow{AB}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} ).
( |\mathbf{w}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} ). 3. ( \cos \theta = \frac{12}{5\sqrt{2} \times \sqrt{3}} = \frac{12}{5\sqrt{6}} ). 4. ( \theta = \cos^{-1}\left(\frac{12}{5\sqrt{6}}\right) \approx 19.5° ).

What we did and why: - The displacement vector is found by subtracting coordinates. - The angle calculation is the same as before, but the context is real-world (drone + wind). - The small angle means the wind is helping the drone move forward.

Common Mistakes

Exam Traps

1-Minute Recap

"Right, listen up—this is your 60-second crash course for vectors in 3D. First, magnitude: square all components, add them, take the square root. Scalar product: multiply matching components and add. Angle between vectors: scalar product divided by magnitudes, then inverse cosine. Negative scalar product? Angle’s obtuse. Zero? They’re perpendicular. Simplify your square roots, and always check if the question wants degrees or radians. You’ve got this—now go smash those exam questions!