Alliance Form 3 P1 Q15 — Vectors: Magnitude of AB

KCSE (Form 3) Form 3 Geometry

Published

The Question

“The position vectors of two points are OA equals 2i plus 3j and OB equals 3i minus 2j. Find the magnitude of the vector AB, giving your answer to one decimal place.”

1

Set up the vector AB

The magnitude of AB is the straight-line distance from A to B, so first you need AB itself as a vector. The rule for the displacement from A to B is to take the position vector of the finishing point and subtract the position vector of the starting point, that is OB minus OA.

AB⃗=OB⃗−OA⃗\vec{AB} = \vec{OB} - \vec{OA}
OA⃗=2i+3j,OB⃗=3i−2j\vec{OA} = 2i + 3j, \quad \vec{OB} = 3i - 2j
2

Subtract the components

Work on the i part and the j part separately. Subtract the i-components and then the j-components, remembering that subtracting a positive 3j lowers the j total. This gives AB as a single vector, which you can also write as a column vector.

AB⃗=(3i−2j)−(2i+3j)\vec{AB} = (3i - 2j) - (2i + 3j)
AB⃗=(3−2)i+(−2−3)j=i−5j\vec{AB} = (3-2)i + (-2-3)j = i - 5j
AB⃗=(1−5)\vec{AB} = \begin{pmatrix} 1 \\ -5 \end{pmatrix}
3

Apply Pythagoras for the magnitude

The magnitude is the length of the arrow, which is the hypotenuse of a right-angled triangle whose horizontal side is the x-component and whose vertical side is the y-component. By Pythagoras, square each component, add them, and take the square root. The sign of each component disappears once you square it.

∣AB⃗∣=x2+y2|\vec{AB}| = \sqrt{x^{2} + y^{2}}
∣AB⃗∣=12+(−5)2=1+25|\vec{AB}| = \sqrt{1^{2} + (-5)^{2}} = \sqrt{1 + 25}
∣AB⃗∣=26≈5.1|\vec{AB}| = \sqrt{26} \approx 5.1

Final Result

The magnitude of vector AB is the square root of 26, which is approximately 5.1 (to one decimal place).

Why this method works

Position vectors tell you where each point sits relative to the origin, but the journey from A to B is the difference between them: you cancel the trip back to the origin from A and then travel out to B, which is why AB equals OB minus OA. Once AB is written in component form, its i-component is the horizontal distance and its j-component is the vertical distance. These two directions are perpendicular, so the actual length of the arrow is the hypotenuse of a right-angled triangle, and Pythagoras gives that hypotenuse exactly. Squaring the components also conveniently removes the negative sign, since a length is always positive.

✓

Check: 1 squared plus 5 squared is 1 plus 25 equals 26, and the square root of 26 is between 5 and 6, closer to 5, giving 5.1 as expected.