AI HL · AHL 4
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Collinear points and parallel lines with vectors
AHL 4.7
In triangle ABC, \(\overrightarrow{AB}=4\mathbf{a}\) and \(\overrightarrow{AC}=3\mathbf{b}\). The point D lies on BC with BD:DC = 2:1. The point E is such that \(\overrightarrow{BE}=2\overrightarrow{AC}\). Write \(\overrightarrow{AE}=t\overrightarrow{AD}\) and find \(t\). This shows that A, D and E lie on a straight line.
Have a go first. Then open the working one line at a time.
\(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-4\mathbf{a}+3\mathbf{b}\)
Go from B to A to C: subtract AB from AC.
\(\overrightarrow{AD}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BC}=\frac{4}{3}\mathbf{a}+2\mathbf{b}=\frac{2}{3}(2\mathbf{a}+3\mathbf{b})\)
BD:DC = 2:1 puts D two thirds of the way along BC.
\(\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+2\overrightarrow{AC}=4\mathbf{a}+6\mathbf{b}=2(2\mathbf{a}+3\mathbf{b})\)
Use BE = 2AC to get from B to E.
\(\overrightarrow{AE}=3\overrightarrow{AD}\). These are parallel and share A, so A, D and E are collinear.
One is a multiple of the other and both start at A. That shows they lie on one line.
In triangle OAB, \(\overrightarrow{OA}=6\mathbf{a}\) and \(\overrightarrow{OB}=4\mathbf{b}\). M is the midpoint of OA and P is the midpoint of AB. Write \(\overrightarrow{MP}=k\overrightarrow{OB}\) and find \(k\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{MA}=\frac{1}{2}\overrightarrow{OA}=3\mathbf{a}\)
M is the midpoint, so OM and MA are each half of OA.
\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-6\mathbf{a}+4\mathbf{b},\quad \overrightarrow{AP}=\frac{1}{2}\overrightarrow{AB}=-3\mathbf{a}+2\mathbf{b}\)
Find AB first, then take half.
\(\overrightarrow{MP}=\overrightarrow{MA}+\overrightarrow{AP}=2\mathbf{b}\)
Go from M to A, then from A to P. The a terms cancel.
\(\overrightarrow{MP}=\frac{1}{2}\overrightarrow{OB}\), so MP is parallel to OB
MP is a multiple of OB, so the lines are parallel.
In triangle OAB, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). The point M is on OB with OM:MB = 3:2, and N is on AB with AN:NB = 1:3. The point E is on OA extended with OA:AE = 7:2. Show that M, N and E are collinear by writing \(\overrightarrow{MN}=k\overrightarrow{NE}\) and finding \(k\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{OM}=\frac{3}{5}\mathbf{b}\)
OM:MB = 3:2 means 3 parts out of 5.
\(\overrightarrow{ON}=\mathbf{a}+\frac{1}{4}(\mathbf{b}-\mathbf{a})=\frac{3}{4}\mathbf{a}+\frac{1}{4}\mathbf{b}\)
Start at A and go one quarter of the way along AB.
\(\overrightarrow{OE}=\frac{9}{7}\mathbf{a}\)
OA:AE = 7:2, so AE is 2/7 of OA and OE = (1 + 2/7)\(\mathbf{a}\).
\(\overrightarrow{MN}=\frac{1}{20}(15\mathbf{a}-7\mathbf{b}),\quad \overrightarrow{NE}=\frac{1}{28}(15\mathbf{a}-7\mathbf{b})\)
Subtract position vectors: MN = ON − OM and NE = OE − ON.
\(\overrightarrow{MN}=\frac{7}{5}\overrightarrow{NE}\)
They are multiples of the same vector and share N, so M, N, E are collinear.
ABCD is a trapezium, \(\overrightarrow{AB}=8\mathbf{a}\), \(\overrightarrow{DC}=4\mathbf{a}\) and \(\overrightarrow{AD}=\mathbf{a}+3\mathbf{b}\). M is the midpoint of AB. Write \(\overrightarrow{DM}=k\overrightarrow{BC}\) and find \(k\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DC}=-3\mathbf{a}+3\mathbf{b}\)
Go from B back to A, along AD, then along DC.
\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}=4\mathbf{a}\)
M is the midpoint of AB.
\(\overrightarrow{DM}=\overrightarrow{DA}+\overrightarrow{AM}=3\mathbf{a}-3\mathbf{b}\)
Go from D to A, then from A to M.
\(\overrightarrow{DM}=-\overrightarrow{BC}=\overrightarrow{CB}\), so DM is parallel to CB.
DM = −BC = CB: same size, opposite direction. The lines are parallel.