ec{AO}+\vec{OB}=\vec{AB}$. ## 2. Points dividing a line in a ratio If P lies on AB and $\text{AP}:\text{PB}=m:n$, then P is $\frac{m}{m+n}$ of the way from A to B. So: $$ \vec{OP}=\vec{OA}+\frac{m}{m+n}\vec{AB} $$ You can also use the weighted form: $$ \vec{OP}=\frac{n}{m+n}\vec{OA}+\frac{m}{m+n}\vec{OB} $$ In the weighted formula, the coefficient of $\vec{OA}$ uses the part near B, and the coefficient of $\vec{OB}$ uses the part near A. In triangle OAB, $\vec{OA}=3a$ and $\vec{OB}=2b$. Point P lies on AB with $\text{AP}:\text{PB}=2:1$. Given that $\vec{OP}=k(3a+4b)$, find $k$. 1. Since AP is 2 out of 3 equal parts, P is $\frac{2}{3}$ of the way from A to B. $$ \vec{OP}=\vec{OA}+\frac{2}{3}\vec{AB} $$ 2. Substitute $\vec{OA}=3a$ and $\vec{AB}=2b-3a$. $$ \begin{aligned} \vec{OP}&=3a+\frac{2}{3}(2b-3a)\\ &=3a+\frac{4}{3}b-2a\\ &=a+\frac{4}{3}b \end{aligned} $$ 3. Factor the expression so it matches the given form. $$ \vec{OP}=\frac{1}{3}(3a+4b) $$ 4. Therefore $k=\frac{1}{3}$. If $\text{AP}:\text{PB}=2:1$, P is $\frac{2}{3}$ of the way from A, not $\frac{1}{3}$ of the way from A. ## 3. Proving points are on the same straight line Three points are collinear if they lie on one straight line. To prove A, B and C are collinear, show that two vectors on the same line are scalar multiples of each other, such as $\vec{AB}=\lambda\vec{AC}$. The cleanest method is: 1. Find position vectors for the points. 2. Choose one point as a common starting point. 3. Form two vectors from that point. 4. Show one is a scalar multiple of the other. In triangle OAB, $\vec{OA}=6a$ and $\vec{OB}=4b$. Point C lies on OA with $\text{OC}:\text{CA}=2:1$. M is the midpoint of AB. Point D lies on OB extended with $\text{OB}:\text{OD}=1:2$. Show that C, M and D are collinear. 1. Find the position vectors of C, M and D. $$ \begin{aligned} \vec{OC}&=\frac{2}{3}(6a)=4a\\ \vec{OM}&=\frac{1}{2}(6a+4b)=3a+2b\\ \vec{OD}&=2(4b)=8b \end{aligned} $$ 2. Use C as the common starting point. $$ \begin{aligned} \vec{CM}&=\vec{OM}-\vec{OC}\\ &=(3a+2b)-4a\\ &=-a+2b \end{aligned} $$ 3. Find the vector from C to D. $$ \begin{aligned} \vec{CD}&=\vec{OD}-\vec{OC}\\ &=8b-4a\\ &=-4a+8b \end{aligned} $$ 4. Compare the two vectors. $$ \vec{CD}=4\vec{CM} $$ 5. Since $\vec{CD}$ is a scalar multiple of $\vec{CM}$, the points C, M and D are collinear. To prove points are on the same straight line, compare vectors that share a point, or clearly state that the parallel vectors pass through a common point. ## 4. Using shape facts For a parallelogram, opposite sides are parallel and equal. If O is one corner and A and B are adjacent corners, the opposite corner C has: $$ \vec{OC}=\vec{OA}+\vec{OB} $$ For a regular hexagon, all sides are equal and all angles are equal. With centre O, opposite vertices have opposite position vectors. PQRSTU is a regular hexagon with centre O. Suppose $\vec{OP}=p$ and $\vec{OQ}=q$. N is the midpoint of QR. Y lies on PQ extended so that $\text{PQ}:\text{QY}=3:2$. Prove that T, N and Y are collinear. 1. Use regular hexagon facts. Since P and Q are adjacent vertices, the next vertex has position vector $\vec{OR}=q-p$, and the opposite vertex has $\vec{OT}=-q$. 2. Find the midpoint N of QR. $$ \begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=-\frac{1}{2}p+q \end{aligned} $$ 3. Since $\text{PQ}:\text{QY}=3:2$, the whole vector from P to Y is $\frac{5}{3}\vec{PQ}$. $$ \begin{aligned} \vec{OY}&=\vec{OP}+\frac{5}{3}\vec{