KCSE 2025 Maths P1 Q21 — Vectors, Column Vectors & Magnitude

KCSE 2025 Form 4 Geometry

Published

The Question

“On a grid the points P and Q have coordinates (3, 3) and (8, 5) respectively, and the vector PQ is shown. (a) Express PQ as a column vector. (b) Two vectors a = (3, -1) and b = (-1, 4) are such that k times a plus h times b equals PQ, where k and h are scalars. Determine the values of k and h. (c) Determine the magnitude of the vector ka.”

1

Find the column vector PQ

The vector from P to Q is the position of Q minus the position of P. Subtract the coordinates entry by entry: the x-parts give the top of the column and the y-parts give the bottom. This tells you how far to move across and up to get from P to Q.

PQ⃗=(85)−(33)=(52)\vec{PQ} = \begin{pmatrix} 8 \\ 5 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}
2

Set up two equations from ka + hb = PQ

Write out k times a plus h times b as a single column vector, then match it row by row against PQ. The top row of the combination must equal the top of PQ, and the bottom row must equal the bottom. This gives two simultaneous equations in the unknown scalars k and h.

k(3−1)+h(−14)=(52)k\begin{pmatrix} 3 \\ -1 \end{pmatrix} + h\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}
3k−h=5(top row)3k - h = 5 \quad \text{(top row)}
−k+4h=2(bottom row)-k + 4h = 2 \quad \text{(bottom row)}
3

Solve the simultaneous equations

Make one unknown the subject of the first equation, then substitute it into the second so only one unknown remains. From the top equation h equals 3k minus 5. Putting that into the bottom equation clears h and leaves a single equation you can solve for k, then back-substitute to get h.

h=3k−5h = 3k - 5
−k+4(3k−5)=2-k + 4(3k - 5) = 2
11k=22⇒k=211k = 22 \Rightarrow k = 2
h=3(2)−5=1h = 3(2) - 5 = 1
4

Find the magnitude of ka

First work out the vector ka by multiplying each entry of a by k, which is 2. Then the magnitude is the length of that vector, found with Pythagoras: square each component, add them, and take the square root. Simplify the surd where you can and give a decimal to check its size.

ka=2(3−1)=(6−2)k\mathbf{a} = 2\begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}
∣ka∣=62+(−2)2=36+4=40|k\mathbf{a}| = \sqrt{6^{2} + (-2)^{2}} = \sqrt{36 + 4} = \sqrt{40}
40=210≈6.32\sqrt{40} = 2\sqrt{10} \approx 6.32

Final Result

PQ is the column vector (5, 2). The scalars are k = 2 and h = 1. The magnitude of ka is the square root of 40, which simplifies to 2 root 10, about 6.32.

Why this method works

A column vector records displacement as change in x over change in y, so subtracting P from Q captures exactly how you travel from one point to the other. Because vectors add component by component, the equation ka + hb = PQ splits cleanly into two ordinary equations, one for each row, which is why simultaneous equations solve it. The magnitude works because a vector's components form the two legs of a right-angled triangle, so its length is the hypotenuse given by Pythagoras.

✓

Substitute k = 2, h = 1 back in: 2(3, -1) + 1(-1, 4) = (6, -2) + (-1, 4) = (5, 2), which matches PQ.