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Year 11 Specialist (Unit 1 & 2) Algebra of vectors

Vector algebra: sums, differences and sections

20 practice questions 0 video lessons Theory + worked examples

Get confident with vector algebra for Year 11 Specialist Mathematics in Queensland (QCAA). Learn to add, subtract and scale vectors in component form, and to find the vector between two points along with its length.

You will also locate the midpoint and any section point that divides a segment in a given ratio, and test whether points are collinear — core skills for the vectors work that runs right through the course.

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Theory

Vector algebra is the set of rules for combining vectors in Year 11 Specialist Mathematics (QCAA, Queensland). You add and subtract vectors component by component, scale them by a number, find the vector between two points, and locate the midpoint or a section point that divides a segment in a given ratio.

A two-dimensional vector is written in component form \(\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the axes. Every operation below is carried out one component at a time.

To add or subtract vectors, add or subtract their matching components. To find a scalar multiple \(k\mathbf{a}\), multiply each component by \(k\); the result is parallel to \(\mathbf{a}\). Two vectors are parallel exactly when one is a scalar multiple of the other, \(\mathbf{a}=\lambda\mathbf{b}\).

The vector between two points \(A\) and \(B\) is \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) (position of the end minus position of the start). Its length \(|\overrightarrow{AB}|\) is the distance from \(A\) to \(B\).

A section point divides a segment \(AB\) in a given ratio. The midpoint is the special case \(1:1\); more generally the point that divides \(AB\) internally in the ratio \(m:n\) is a weighted average of the two endpoints. Three points are collinear when the vectors joining them are parallel.

Triangle law of vector addition Vector a runs from the origin to the point four across and one up. Vector b starts at the end of a and runs one across and three up. The resultant a plus b runs straight from the origin to five across and four up, closing the triangle. x y a b a + b
Triangle law: place \(\mathbf{b}\) tip-to-tail after \(\mathbf{a}\); the resultant \(\mathbf{a}+\mathbf{b}\) closes the triangle.
A section point dividing a segment in a ratio The segment from A at minus two, minus one to B at ten, five is drawn. Point P at six, three lies on it, dividing AB internally in the ratio two to one, so the part from A to P is twice the part from P to B. x y A(-2,-1) B(10,5) P(6,3) 2 1
Section point: \(P(6,3)\) divides \(AB\) internally in the ratio \(2:1\), so \(AP\) is twice \(PB\).

For \(\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}\) and \(\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}\), the sum, difference and scalar multiple are:

\[ \mathbf{a}\pm\mathbf{b}=(a_1\pm b_1)\mathbf{i}+(a_2\pm b_2)\mathbf{j},\qquad k\mathbf{a}=ka_1\mathbf{i}+ka_2\mathbf{j} \]
a±b=(a1±b1)i+(a2±b2)j

For points \(A(a_1,a_2)\) and \(B(b_1,b_2)\), the joining vector, its length and the midpoint are:

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a},\qquad |\overrightarrow{AB}|=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2},\qquad M=\left(\dfrac{a_1+b_1}{2},\ \dfrac{a_2+b_2}{2}\right) \]
|AB|=(b1a1)2+(b2a2)2

The point that divides \(AB\) internally in the ratio \(m:n\) is:

\[ P=\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n} \]
P=na+mbm+n
Weights sit on the far endpoint. For ratio \(m:n\) the weight \(m\) (the part nearest \(B\)) multiplies \(\mathbf{a}\)’s opposite endpoint \(\mathbf{b}\), and \(n\) multiplies \(\mathbf{a}\). The midpoint is just \(m:n=1:1\), giving \(\tfrac12(\mathbf{a}+\mathbf{b})\).

How to work with vectors and section points

  1. Write each vector in components: put every vector as \(a_1\mathbf{i}+a_2\mathbf{j}\) (or as coordinates for a point), so you can operate one component at a time.
  2. Add, subtract or scale component-wise: combine the \(\mathbf{i}\) parts and the \(\mathbf{j}\) parts separately; multiply every component by the scalar.
  3. Between two points, subtract positions: use \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), then \(|\overrightarrow{AB}|=\sqrt{(\Delta x)^2+(\Delta y)^2}\) for its length.
  4. For a section point, weight the endpoints: for ratio \(m:n\) form \(\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n}\); for the midpoint just average the coordinates.
Example 1 — Sum and scalar multiple
Given \(\mathbf{a}=3\mathbf{i}+2\mathbf{j}\) and \(\mathbf{b}=5\mathbf{i}-4\mathbf{j}\), find \(\mathbf{a}+\mathbf{b}\) and \(2\mathbf{a}-\mathbf{b}\).
Solution

Add the matching components for the sum:

\(\mathbf{a}+\mathbf{b}\)\(=\)\((3+5)\mathbf{i}+(2+(-4))\mathbf{j}\)
\(=\)\(8\mathbf{i}-2\mathbf{j}\)

Scale \(\mathbf{a}\) first, then subtract \(\mathbf{b}\):

\(2\mathbf{a}\)\(=\)\(6\mathbf{i}+4\mathbf{j}\)
\(2\mathbf{a}-\mathbf{b}\)\(=\)\((6-5)\mathbf{i}+(4-(-4))\mathbf{j}\)
\(=\)\(\mathbf{i}+8\mathbf{j}\)

\(\mathbf{a}+\mathbf{b}=8\mathbf{i}-2\mathbf{j}\) and \(2\mathbf{a}-\mathbf{b}=\mathbf{i}+8\mathbf{j}\).

Example 2 — Vector between points and its length
Points \(A(2,-1)\) and \(B(7,11)\) are given. Find \(\overrightarrow{AB}\) and \(|\overrightarrow{AB}|\).
Solution

Subtract the position of the start from the position of the end:

\(\overrightarrow{AB}\)\(=\)\(\mathbf{b}-\mathbf{a}\)
\(=\)\((7-2)\mathbf{i}+(11-(-1))\mathbf{j}\)
\(=\)\(5\mathbf{i}+12\mathbf{j}\)

Apply the length formula to that vector:

\(|\overrightarrow{AB}|\)\(=\)\(\sqrt{5^2+12^2}\)
\(=\)\(\sqrt{25+144}\)
\(=\)\(\sqrt{169}\)
\(=\)\(13\)

\(\overrightarrow{AB}=5\mathbf{i}+12\mathbf{j}\), of length \(|\overrightarrow{AB}|=13\).

Example 3 — Midpoint
Find the midpoint \(M\) of the segment joining \(C(-3,4)\) and \(D(5,-2)\).
Solution

The midpoint is the average of the two endpoints’ coordinates:

\(M\)\(=\)\(\left(\dfrac{-3+5}{2},\ \dfrac{4+(-2)}{2}\right)\)
\(=\)\(\left(\dfrac{2}{2},\ \dfrac{2}{2}\right)\)
\(=\)\((1,1)\)

The midpoint is \(M(1,1)\).

Example 4 — Section point in a ratio
Point \(P\) divides the segment from \(A(1,2)\) to \(B(11,7)\) internally in the ratio \(3:2\). Find \(P\).
Solution

Use \(P=\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n}\) with \(m:n=3:2\), so the far weight \(3\) multiplies \(\mathbf{b}\):

\(P\)\(=\)\(\dfrac{2(1,2)+3(11,7)}{3+2}\)
\(=\)\(\left(\dfrac{2+33}{5},\ \dfrac{4+21}{5}\right)\)
\(=\)\(\left(\dfrac{35}{5},\ \dfrac{25}{5}\right)\)
\(=\)\((7,5)\)

\(P=(7,5)\).

A section point dividing a segment in a ratio The segment from A at minus two, minus one to B at ten, five is drawn. Point P at six, three lies on it, dividing AB internally in the ratio two to one, so the part from A to P is twice the part from P to B. x y A(-2,-1) B(10,5) P(6,3) 2 1

Common pitfalls

Subtracting the points the wrong way round. The vector from \(A\) to \(B\) is \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) (end minus start). Writing \(\mathbf{a}-\mathbf{b}\) gives \(\overrightarrow{BA}\), which points the opposite way.
Mishandling a double negative. When a component is negative, keep the signs explicit: \(2-(-4)=6\), not \(2-4=-2\). Most sign errors happen at this step.
Putting the section weights on the wrong endpoint. For ratio \(m:n\) the formula is \(\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n}\): the first number \(m\) multiplies \(\mathbf{b}\), not \(\mathbf{a}\). Swapping them lands you on the wrong side of the midpoint.
Forgetting to take the square root for magnitude. \(|\mathbf{v}|=\sqrt{v_1^2+v_2^2}\); leaving the answer as \(v_1^2+v_2^2\) gives the squared length, not the length.

Frequently asked questions

How do you add or subtract two vectors in component form?

Add or subtract the matching components separately. For \(\mathbf{a}=3\mathbf{i}+2\mathbf{j}\) and \(\mathbf{b}=5\mathbf{i}-4\mathbf{j}\), \(\mathbf{a}+\mathbf{b}=8\mathbf{i}-2\mathbf{j}\) and \(\mathbf{a}-\mathbf{b}=-2\mathbf{i}+6\mathbf{j}\).

How do you find the vector between two points?

Subtract the position of the start point from the position of the end point: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\). For \(A(2,-1)\) and \(B(7,11)\) this gives \(5\mathbf{i}+12\mathbf{j}\).

What is the section formula for dividing a segment in a ratio?

A point dividing \(AB\) internally in the ratio \(m:n\) is \(P=\dfrac{n\mathbf{a}+m\mathbf{b}}{m+n}\). The midpoint is the case \(m:n=1:1\), giving \(\tfrac12(\mathbf{a}+\mathbf{b})\).

How do you find the midpoint of two points?

Average the coordinates: \(M=\left(\dfrac{a_1+b_1}{2},\ \dfrac{a_2+b_2}{2}\right)\). For \(C(-3,4)\) and \(D(5,-2)\) the midpoint is \((1,1)\).

How do you show that three points are collinear using vectors?

Form two vectors joining the points, such as \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). If one is a scalar multiple of the other they are parallel, and since they share a point the three points lie on one straight line.

How do you find the magnitude of a vector?

Use \(|\mathbf{v}|=\sqrt{v_1^2+v_2^2}\). For \(\mathbf{v}=5\mathbf{i}+12\mathbf{j}\), \(|\mathbf{v}|=\sqrt{25+144}=\sqrt{169}=13\).