Two-dimensional vector arithmetic and magnitude — 0580
Operate on corresponding components and use Pythagoras for length.
Extended · Syllabus E7.2, E7.3
Notes and key methods
A two-dimensional vector has horizontal and vertical components. Add, subtract and multiply by a scalar component by component. The displacement AB is position(B)−position(A), so order matters.
The magnitude of (a,b) is \(\displaystyle \sqrt{a^{2}+b^{2}}\), a nonnegative scalar. Parallel nonzero vectors are scalar multiples. These 0580 exercises use two dimensions and do not require scalar products, cross products or three-dimensional vector equations.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Two-dimensional vector arithmetic and magnitude
Vectors \(\displaystyle u=\)\(\left[\begin{matrix}12\\16\end{matrix}\right]\), \(\displaystyle v=\)\(\left[\begin{matrix}8\\0\end{matrix}\right]\) and \(\displaystyle w=2u-5v\).
Find the horizontal component of w.
Find the vertical component of w.
Find the magnitude of u.
Find the magnitude of 5u.
Show the reasoning
(a)
Combine the horizontal components with their signed scalar multipliers.
Twice 12 minus 5 times 8 is \(-16\).
Answer: \(-16\).
(b)
Combine the vertical components, preserving a negative component if present.
Twice 16 minus 5 times 0 is \(32\).
Answer: \(32\).
(c)
Use Pythagoras on the two components.
\(\displaystyle \sqrt{144+256}=20\).
Answer: \(20\).
(d)
Scale the vector magnitude by the absolute scalar multiplier.
A positive scalar 5 multiplies the magnitude 20 to give 100.
Answer: \(100\).
Example 2 · Two-dimensional vector arithmetic and magnitude
Vectors \(\displaystyle u=\)\(\left[\begin{matrix}15\\20\end{matrix}\right]\), \(\displaystyle v=\)\(\left[\begin{matrix}2\\1\end{matrix}\right]\) and \(\displaystyle w=2u-2v\).
Find the horizontal component of w.
Find the vertical component of w.
Find the magnitude of u.
Find the magnitude of 2u.
Show the reasoning
(a)
Combine the horizontal components with their signed scalar multipliers.
Twice 15 minus 2 times 2 is \(26\).
Answer: \(26\).
(b)
Combine the vertical components, preserving a negative component if present.
Twice 20 minus 2 times 1 is \(38\).
Answer: \(38\).
(c)
Use Pythagoras on the two components.
\(\displaystyle \sqrt{225+400}=25\).
Answer: \(25\).
(d)
Scale the vector magnitude by the absolute scalar multiplier.
A positive scalar 2 multiplies the magnitude 25 to give 50.
Answer: \(50\).
Example 3 · Two-dimensional vector arithmetic and magnitude
Vectors \(\displaystyle u=\)\(\left[\begin{matrix}18\\24\end{matrix}\right]\), \(\displaystyle v=\)\(\left[\begin{matrix}3\\2\end{matrix}\right]\) and \(\displaystyle w=2u-3v\).
Find the horizontal component of w.
Find the vertical component of w.
Find the magnitude of u.
Find the magnitude of 3u.
Show the reasoning
(a)
Combine the horizontal components with their signed scalar multipliers.
Twice 18 minus 3 times 3 is \(27\).
Answer: \(27\).
(b)
Combine the vertical components, preserving a negative component if present.
Twice 24 minus 3 times 2 is \(42\).
Answer: \(42\).
(c)
Use Pythagoras on the two components.
\(\displaystyle \sqrt{324+576}=30\).
Answer: \(30\).
(d)
Scale the vector magnitude by the absolute scalar multiplier.
A positive scalar 3 multiplies the magnitude 30 to give 90.
Answer: \(90\).
Common mistakes
Keep intermediate values exact until the final step.