Cambridge O Level Additional Mathematics · Syllabus 4037 · Vectors in Two Dimensions
Scalar Multiple
What is Scalar Multiple?
The vector obtained by multiplying every component of a vector by the same number k, written k a. Its magnitude is the modulus of k times the magnitude of the original vector, so the vector is stretched when the modulus of k exceeds 1 and compressed when it is less than 1. Its direction is unchanged when k is positive and exactly reversed when k is negative; multiplying by zero gives the zero vector. Two non-zero vectors are parallel precisely when one is a scalar multiple of the other, which is why the scalar multiple is the standard tool for proving that lines are parallel and that points are collinear.
This definition is part of the Vectors in Two Dimensions chapter in Cambridge O Level Additional Mathematics.
Scalar Multiple in context
A unit vector has magnitude exactly \(1\), found by dividing a vector by its own magnitude, \(\widehat{\mathbf a}=\mathbf a/|\mathbf a|\); it is undefined for the zero vector, since division by zero has no meaning. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, and points \(A\), \(B\), \(C\) are collinear only when that scalar multiple also shares a common point, since parallel segments can still lie on different lines. Two particles moving with constant velocity collide only when their \(x\)- and \(y\)-component equations give the same time \(t\), not merely when their paths cross.
Common mistakes with Scalar Multiple
- Concluding that three points are collinear because they look collinear on the diagram. Why it fails Diagrams in questions are not necessarily drawn to scale, so their proportions carry no guarantee. Three points can look collinear and be a long way from it; the human eye cannot resolve a gradient difference of a few per cent. Fix Compute two displacement vectors and show one is a scalar multiple of the other. That is the proof, and no amount of drawing substitutes for it.
- Proving \(\overrightarrow{AB}=k\,\overrightarrow{CD}\) for four distinct points and concluding that \(A\), \(B\), \(C\), \(D\) are collinear. Why it fails Nothing in that statement forces the two segments onto the same line; parallel segments sitting on different lines satisfy it perfectly well. The shared point is the missing ingredient, and it cannot be inferred from the scalar multiple. Fix Choose displacements that already share a letter — \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\), or \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) — and then say in words which point is common to both.
- Reaching for a dot product, a cross product or a third component. Why it fails None of these is in Additional Mathematics 4037, so a method that depends on one cannot be what a question is testing. More practically, a solution built on an unavailable tool tends to bypass exactly the reasoning the question was set to draw out — the component equations, the scalar multiple, the shared time. Fix If your method needs \(\mathbf a\cdot\mathbf b\) or a \(\mathbf k\) component, stop and re-read the question: there is a two-dimensional route through it using only the operations in this chapter. Perpendicularity, when it appears at this level, is set up geometrically rather than by a dot product.
Questions students ask about Scalar Multiple
How do you prove that two vectors are parallel, or that three points are collinear?
Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, using the same scalar on both rows. Three points \(A\), \(B\), \(C\) are collinear when a displacement such as \(\overrightarrow{AB}\) is a scalar multiple of \(\overrightarrow{BC}\) and the two share the point \(B\); the scalar multiple alone only proves the segments are parallel, since parallel segments can lie on different lines. A diagram that looks straight is never sufficient evidence, since exam diagrams are not drawn to scale.