{5}$. If $AP:PB=2:3$, do not use $\frac{3}{5}$ from A. The 3 parts are from P to B, not from A to P. ## 3. Proving three points are on the same straight line Three points on one straight line are called **collinear**. To prove collinearity, you usually show that two vectors are scalar multiples of each other. A **scalar** is an ordinary number. If one vector is a scalar multiple of another, such as $\vec{AD}=\lambda\vec{AE}$, then the two vectors are parallel and lie along the same line. The target is to compare vectors that start from the same point. ![Collinearity vector proof diagram](https://assets.mathsgenie.co.uk/notes/db551ac3-ec65-4cac-9624-accb12bb26db.jpeg) To prove A, D and E are collinear, try to show $\vec{AD}=\lambda\vec{AE}$, or $\vec{AE}=\lambda\vec{AD}$, for some number $\lambda$. OABC is a parallelogram with adjacent sides $\vec{OA}=3a$ and $\vec{OB}=3b$. Point D lies on OC so that $OD:DC=2:1$. Point E is the midpoint of BC. Show that A, D and E lie on one straight line. 1. In a parallelogram, the diagonal position vector is found by adding the adjacent sides. $$ \vec{OC}=3a+3b $$ 2. Since $OD:DC=2:1$, D is $\frac{2}{3}$ of the way from O to C. $$ \vec{OD}=\frac{2}{3}(3a+3b)=2a+2b $$ 3. E is the midpoint of B and C. $$ \begin{aligned} \vec{OE}&=\frac{1}{2}(\vec{OB}+\vec{OC})\\ &=\frac{1}{2}(3b+3a+3b)\\ &=\frac{3}{2}a+3b \end{aligned} $$ 4. Now compare vectors starting from A. $$ \begin{aligned} \vec{AD}&=\vec{OD}-\vec{OA}\\ &=(2a+2b)-3a\\ &=-a+2b \end{aligned} $$ 5. Also find $\vec{AE}$. $$ \begin{aligned} \vec{AE}&=\vec{OE}-\vec{OA}\\ &=\left(\frac{3}{2}a+3b\right)-3a\\ &=-\frac{3}{2}a+3b\\ &=\frac{3}{2}(-a+2b) \end{aligned} $$ 6. Since $\vec{AE}=\frac{3}{2}\vec{AD}$, the vectors are scalar multiples. Therefore A, D and E are collinear. To prove A, D and E are collinear, do not compare $\vec{OD}$ and $\vec{OE}$. That would only tell you about a line through O. Use vectors such as $\vec{AD}$ and $\vec{AE}$. ## 4. Regular hexagon vector facts In a regular hexagon, the centre-to-vertex vectors are very useful. If two adjacent position vectors are $p$ and $q$, the next vertex can often be written using $q-p$. The diagram shows the common pattern. ![Regular hexagon vector positions](https://assets.mathsgenie.co.uk/notes/cec60095-bdd3-4eb9-8cdf-78dfce7845ba.jpeg) For a regular hexagon PQRSTU with centre O: - If $\vec{OP}=p$ and $\vec{OQ}=q$, then $\vec{OR}=q-p$. - Opposite vertices have opposite position vectors, so $\vec{OT}=-q$ and $\vec{OS}=-p$. PQRSTU is a regular hexagon with centre O. Let $\vec{OP}=p$ and $\vec{OQ}=q$. Point N is the midpoint of QR. Point Y lies on PQ extended beyond Q, with $PQ:QY=3:2$. Prove that T, N and Y are collinear. 1. Use the regular hexagon fact for the next vertex. $$ \vec{OR}=q-p $$ 2. N is the midpoint of Q and R. $$ \begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=q-\frac{1}{2}p \end{aligned} $$ 3. Since Y is beyond Q and $PQ:QY=3:2$, use $\vec{QY}=\frac{2}{3}\vec{PQ}$. $$ \begin{aligned} \vec{OY}&=\vec{OQ}+\vec{QY}\\ &=q+\frac{2}{3}(q-p)\\ &=\frac{5}{3}q-\frac{2}{3}p \end{aligned} $$ 4. The vertex T is opposite Q, so $\vec{OT}=-q$. 5. Compare vectors from T. $$ \begin{aligned} \vec{TN}&=\vec{ON}-\vec{OT}\\ &=\left(q-\frac{1}{2}p\right)-(-q)\\ &=2q-\frac{1}{2}p \end{aligned} $$ 6. Now find $\vec{TY}$. $$ \begin{aligned} \vec{TY}&=\vec{OY}-\vec{OT}\\ &=\left(\frac{5}{3}q-\frac{2}{3