3.4.1.1
Two perpendicular vectors are added using Pythagoras and trigonometry
Scalars & Vectors — AQA A-Level Physics
- Link the two vectors head-to-tail to form a right-angled triangle.
- The resultant is the hypotenuse from the tail of the first vector to the head of the second.
- Find the magnitude using Pythagoras: $R = \sqrt{a^2 + b^2}$.
- Find the direction using trigonometry: $\tan(\theta) = opposite / adjacent.$
$$R = \sqrt{a^2 + b^2}$$
- $R$: magnitude of resultant vector
- $a$: magnitude of first vector
- $b$: magnitude of second vector (perpendicular to a)
Worked Example
A hiker walks 6 km due east and 10 km due north. Calculate the magnitude and direction of their displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). from the horizontal.
Show Solution
1
Sketch the vector triangle
The two legs are perpendicular, forming a right-angled triangle with the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). as hypotenuse.
2
Calculate magnitude using Pythagoras
$$R = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} = 11.7 \text{ km}$$
3
Calculate direction using trigonometry
$$\tan \theta = \frac{10}{6}$$
$$\theta = \tan^{-1}\left(\frac{10}{6}\right) = 59°$$
The displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). is 59 degrees from the horizontal (east).
Answer
$Displacement = 11.7 km at 59 degrees from the horizontal.$
Examiner Tips and Tricks
- Pythagoras and trigonometry appear in nearly every vector question.
- Make sure you are fully confident with both before the exam.
Related:Kinematics