Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — combining vectors and finding a magnitude
Given a=3i+4j and b=i−2j, find a+2b and ∣a∣.
- 12b=2i−4jA scalar multiplies both components, not just the first.
- 2a+2b=(3+2)i+(4−4)jAdd the i parts and the j parts separately.
- 3=5iThe j component is zero, so the result points straight along the x-axis.
- 4∣a∣=32+42=25=5Pythagoras on the components — square, add, root.
Example 2 — the vector joining two points, and its unit vector
The points are A(1,2) and B(5,5). Find AB, its magnitude, and the unit vector in the direction of AB.
- 1AB=OB−OADestination minus start. Getting this the wrong way round reverses the vector.
- 2=(5−1)i+(5−2)j=4i+3jSubtract the coordinates component by component.
- 3∣AB∣=16+9=5A 3-4-5 triangle.
- 4u^=51(4i+3j)=0.8i+0.6jDivide by the magnitude. Check: 0.64+0.36=1 ✓
Example 3 — using the scalar product to find an unknown
Given a=2i+3j and b=6i+kj, find the value of k for which a and b are perpendicular.
- 1Perpendicular means a⋅b=0.This is the whole reason the scalar product is on the syllabus.
- 2a⋅b=(2)(6)+(3)(k)=12+3kMultiply matching components and add.
- 312+3k=0⇒k=−4Divide by 3 at the end.
- 4Check: b=6i−4j and 12−12=0 ✓Always substitute back — it costs five seconds.