Column Vectors, Vector Geometry & Collinear Vector Proofs

1

Column Vectors & Scalar Multiplication

A column vector (x / y) represents a horizontal movement of x units (positive = right, negative = left) and a vertical movement of y units (positive = up, negative = down). To add or subtract column vectors, add or subtract the top components and bottom components separately. To multiply by a scalar k, multiply both components by k.

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2

Triangle Law of Vector Addition ($ec{AB} = ec{AO} + ec{OB}$)

To find the vector from any point X to any point Y, chain together known vectors along a path from X to Y: $ec{XY} = ec{XO} + ec{OY}$. Remember that reversing the direction of a vector changes its sign: $ec{XO} = -ec{OX}$.

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3

Midpoints & Ratio Division of a Line Segment

1. If M is the midpoint of AB, then $ec{AM} = rac{1}{2}ec{AB}$. 2. If a point P divides the line segment AB in the ratio m : n (so AP : PB = m : n), then the total number of parts is (m + n), so $ec{AP} = rac{m}{m+n}ec{AB}$ and $ec{PB} = rac{n}{m+n}ec{AB}$.

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4

Parallel Vectors Condition ($mathbf{u} = kmathbf{v}$)

Two non-zero vectors **u** and **v** are parallel if and only if one vector can be written as a scalar multiple of the other (**u** = k**v**, where k is a number). Always factorise out the coefficients of **a** and **b** to reveal the identical bracket (e.g., $6mathbf{a} - 9mathbf{b} = 3(2mathbf{a} - 3mathbf{b})$ is parallel to $4mathbf{a} - 6mathbf{b} = 2(2mathbf{a} - 3mathbf{b})$).

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5

Collinear Points Proof (Higher Tier)

Three points P, Q, and R are collinear if they lie on the exact same straight line. To prove P, Q, and R are collinear: 1. Work out the vectors $ec{PQ}$ and $ec{QR}$ (or $ec{PR}$) in terms of **a** and **b** and simplify/factorise both. 2. Show that $ec{PQ} = kec{QR}$, which proves PQ is parallel to QR. 3. Write the mandatory conclusion: 'Since $ec{PQ} = kec{QR}$, PQ is parallel to QR, and since they share the common point Q, P, Q, and R are collinear.'

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6

Equating Coefficients of Non-Parallel Vectors (Higher Tier)

If **a** and **b** are two non-parallel vectors and $pmathbf{a} + qmathbf{b} = rmathbf{a} + smathbf{b}$, then the coefficients of **a** must be equal ($p = r$) and the coefficients of **b** must be equal ($q = s$). This technique solves Grade 9 intersection ratio problems.

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1. Q1

   In triangle OAB, $\vec{OA} = 2\mathbf{a}$ and $\vec{OB} = 3\mathbf{b}$. P is the point on AB such that AP : PB = 2 : 3. Find $\vec{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ in its simplest form. (Higher Tier) [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Find the vector $\vec{AB}$ first by going from A to O to B: $\vec{AB} = \vec{AO} + \vec{OB} = -2\mathbf{a} + 3\mathbf{b} = 3\mathbf{b} - 2\mathbf{a}$ [1 mark]. 2. Since AP : PB = 2 : 3, there are 2 + 3 = 5 parts in total, so $\vec{AP} = \frac{2}{5}\vec{AB} = \frac{2}{5}(3\mathbf{b} - 2\mathbf{a})$ [1 mark]. 3. Write a vector route for $\vec{OP}$ via A: $\vec{OP} = \vec{OA} + \vec{AP} = 2\mathbf{a} + \frac{2}{5}(3\mathbf{b} - 2\mathbf{a})$ [1 mark]. 4. Expand and collect like terms: $\vec{OP} = 2\mathbf{a} + \frac{6}{5}\mathbf{b} - \frac{4}{5}\mathbf{a} = \frac{6}{5}\mathbf{a} + \frac{6}{5}\mathbf{b}$ (or $\frac{6}{5}(\mathbf{a} + \mathbf{b})$) [1 mark].

2. Q2

   OABC is a parallelogram where $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$. P is the midpoint of the side AB. The line OA is extended to a point N such that OA = AN (so $\vec{ON} = 2\mathbf{a}$). Prove using vectors that the points C, P, and N are collinear. (Higher Tier) [5 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. In parallelogram OABC, opposite sides are equal and parallel, so $\vec{CB} = \vec{OA} = \mathbf{a}$ and $\vec{AB} = \vec{OC} = \mathbf{c}$ [1 mark]. 2. Since P is the midpoint of AB, $\vec{BP} = -\frac{1}{2}\mathbf{c}$ (and $\vec{AP} = \frac{1}{2}\mathbf{c}$). Find vector $\vec{CP}$: $\vec{CP} = \vec{CB} + \vec{BP} = \mathbf{a} - \frac{1}{2}\mathbf{c}$ [1 mark]. 3. Find vector $\vec{CN}$ by going from C to O to N: $\vec{CN} = \vec{CO} + \vec{ON} = -\mathbf{c} + 2\mathbf{a} = 2\mathbf{a} - \mathbf{c}$ [1 mark]. 4. Factorise $\vec{CN}$ to compare with $\vec{CP}$: $\vec{CN} = 2(\mathbf{a} - \frac{1}{2}\mathbf{c}) = 2\vec{CP}$ [1 mark]. 5. Conclusion: Since $\vec{CN} = 2\vec{CP}$, $\vec{CN}$ is a scalar multiple of $\vec{CP}$ so CN and CP are parallel; since they also share the common point C, the points C, P, and N are collinear [1 mark].

3. Q3

   Given the column vectors $\mathbf{p} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$, $\mathbf{q} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$, and $\mathbf{r} = \begin{pmatrix} 14 \\ -21 \end{pmatrix}$, find the scalars $h$ and $k$ such that $h\mathbf{p} + k\mathbf{q} = \mathbf{r}$, and calculate the exact magnitude of vector $\mathbf{p}$. [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Set up two simultaneous equations from the top (x) and bottom (y) components of $h\mathbf{p} + k\mathbf{q} = \mathbf{r}$: $4h - 2k = 14$ and $-3h + 5k = -21$ [1 mark]. 2. Multiply the first equation by 5 ($20h - 10k = 70$) and the second by 2 ($-6h + 10k = -42$), then add them together to eliminate $k$: $14h = 28 \rightarrow h = 2$ [1 mark]. 3. Substitute $h = 2$ into $4h - 2k = 14$: $8 - 2k = 14 \rightarrow -2k = 6 \rightarrow k = -3$ [1 mark]. 4. Magnitude of $\mathbf{p} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ [1 mark].

4. Q4

   In triangle OAB, $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. M is the midpoint of OA. P is a point on AB such that $\vec{AP} = k\vec{AB}$, where $k$ is a scalar. Given that $\vec{MP} = \frac{1}{6}\mathbf{a} + \frac{1}{3}\mathbf{b}$, find the value of $k$ and hence state the ratio AP : PB. (Higher Tier) [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Since M is the midpoint of OA, $\vec{MA} = \frac{1}{2}\mathbf{a}$; also $\vec{AB} = \mathbf{b} - \mathbf{a}$ [1 mark]. 2. Write an expression for $\vec{MP}$ in terms of $k$, $\mathbf{a}$, and $\mathbf{b}$: $\vec{MP} = \vec{MA} + \vec{AP} = \frac{1}{2}\mathbf{a} + k(\mathbf{b} - \mathbf{a}) = (\frac{1}{2} - k)\mathbf{a} + k\mathbf{b}$ [1 mark]. 3. Equate the coefficients of $\mathbf{a}$ and $\mathbf{b}$ with the given vector $\vec{MP} = \frac{1}{6}\mathbf{a} + \frac{1}{3}\mathbf{b}$: comparing coefficients of $\mathbf{b}$ gives $k = \frac{1}{3}$ (and checking $\mathbf{a}$: $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$) [1 mark]. 4. Since $\vec{AP} = \frac{1}{3}\vec{AB}$, AP is 1 part out of 3 and PB is 2 parts out of 3, so the ratio AP : PB = 1 : 2 [1 mark].

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