Collinear points and parallel lines with vectors Problem set A Worked solutions

In triangle ABC, \(\overrightarrow{AB}=2\mathbf{a}\) and \(\overrightarrow{AC}=5\mathbf{b}\). The point D lies on BC with BD:DC = 3:2. The point E is such that \(\overrightarrow{BE}=\frac{3}{2}\overrightarrow{AC}\). Show that A, D and E lie on a straight line.

• \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-2\mathbf{a}+5\mathbf{b}\)
• \(\overrightarrow{AD}=\overrightarrow{AB}+\frac{3}{5}\overrightarrow{BC}=\frac{4}{5}\mathbf{a}+3\mathbf{b}=\frac{1}{5}(4\mathbf{a}+15\mathbf{b})\)
• \(\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}=2\mathbf{a}+\frac{15}{2}\mathbf{b}=\frac{1}{2}(4\mathbf{a}+15\mathbf{b})\)
• \(\overrightarrow{AE}=\frac{5}{2}\overrightarrow{AD}\). These are parallel and share A, so A, D and E are collinear.

In triangle OAB, \(\overrightarrow{OA}=4\mathbf{a}\) and \(\overrightarrow{OB}=6\mathbf{b}\). M is the midpoint of OB and N is the midpoint of AB. Show that MN is parallel to OA.

• \(\overrightarrow{MB}=\frac{1}{2}\overrightarrow{OB}=3\mathbf{b}\)
• \(\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=4\mathbf{a}-6\mathbf{b},\quad \overrightarrow{BN}=\frac{1}{2}\overrightarrow{BA}=2\mathbf{a}-3\mathbf{b}\)
• \(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=2\mathbf{a}\)
• \(\overrightarrow{MN}=\frac{1}{2}\overrightarrow{OA}\), so MN is parallel to OA

Show that the points A(1, 2, −1), B(4, −1, 5) and C(7, −4, 11) are collinear.

• \(\overrightarrow{AB}=\begin{pmatrix}4\\-1\\5\end{pmatrix}-\begin{pmatrix}1\\2\\-1\end{pmatrix}=\begin{pmatrix}3\\-3\\6\end{pmatrix}\)
• \(\overrightarrow{AC}=\begin{pmatrix}7\\-4\\11\end{pmatrix}-\begin{pmatrix}1\\2\\-1\end{pmatrix}=\begin{pmatrix}6\\-6\\12\end{pmatrix}\)
• \(\overrightarrow{AC}=2\overrightarrow{AB}\). The vectors are parallel and share A, so the points are collinear.

A triangular sail ABC has a frame, \(\overrightarrow{AB}=3\mathbf{a}\) and \(\overrightarrow{AC}=4\mathbf{b}\). The point D lies on BC with BD:DC = 1:3. The point E is such that \(\overrightarrow{BE}=\frac{1}{3}\overrightarrow{AC}\). Show that A, D and E lie on a straight line.

• \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-3\mathbf{a}+4\mathbf{b}\)
• \(\overrightarrow{AD}=\overrightarrow{AB}+\frac{1}{4}\overrightarrow{BC}=\frac{9}{4}\mathbf{a}+\mathbf{b}=\frac{1}{4}(9\mathbf{a}+4\mathbf{b})\)
• \(\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=3\mathbf{a}+\frac{4}{3}\mathbf{b}=\frac{1}{3}(9\mathbf{a}+4\mathbf{b})\)
• \(\overrightarrow{AE}=\frac{4}{3}\overrightarrow{AD}\). These are parallel and share A, so A, D and E are collinear.

A garden bed ABCD is a trapezium (lengths in metres), \(\overrightarrow{AB}=10\mathbf{a}\), \(\overrightarrow{DC}=5\mathbf{a}\) and \(\overrightarrow{AD}=2\mathbf{a}+4\mathbf{b}\). M is the midpoint of AB. Show that DM is parallel to CB.

• \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DC}=-3\mathbf{a}+4\mathbf{b}\)
• \(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}=5\mathbf{a}\)
• \(\overrightarrow{DM}=\overrightarrow{DA}+\overrightarrow{AM}=3\mathbf{a}-4\mathbf{b}\)
• \(\overrightarrow{DM}=-\overrightarrow{BC}=\overrightarrow{CB}\), so DM is parallel to CB.

A drone is seen at three positions P(2, −1, 4), Q(5, 2, −2) and R(11, 8, −14) (coordinates in metres). Show that the drone moved in a straight line.

• \(\overrightarrow{PQ}=\begin{pmatrix}5\\2\\-2\end{pmatrix}-\begin{pmatrix}2\\-1\\4\end{pmatrix}=\begin{pmatrix}3\\3\\-6\end{pmatrix}\)
• \(\overrightarrow{PR}=\begin{pmatrix}11\\8\\-14\end{pmatrix}-\begin{pmatrix}2\\-1\\4\end{pmatrix}=\begin{pmatrix}9\\9\\-18\end{pmatrix}\)
• \(\overrightarrow{PR}=3\overrightarrow{PQ}\). The vectors are parallel and share P, so the points are collinear.

In triangle ABC, \(\overrightarrow{AB}=5\mathbf{a}\) and \(\overrightarrow{AC}=2\mathbf{b}\). The point D lies on BC with BD:DC = 2:3. The point E is such that \(\overrightarrow{BE}=k\overrightarrow{AC}\). Find the value of \(k\) for which A, D and E lie on a straight line.

• \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-5\mathbf{a}+2\mathbf{b}\)
• \(\overrightarrow{AD}=\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BC}=3\mathbf{a}+\frac{4}{5}\mathbf{b}=\frac{1}{5}(15\mathbf{a}+4\mathbf{b})\)
• \(\overrightarrow{AE}=\overrightarrow{AB}+k\overrightarrow{AC}=5\mathbf{a}+2k\mathbf{b}\)
• For a straight line \(\overrightarrow{AE}=t\overrightarrow{AD}\). Compare \(\mathbf{a}\): \(5=t\times 3\), so \(t=\frac{5}{3}\)
• Compare \(\mathbf{b}\): \(2k=\frac{5}{3}\times \frac{4}{5}\), so \(k=\frac{2}{3}\)

WXYZ is a quadrilateral, \(\overrightarrow{WX}=8\mathbf{a}\), \(\overrightarrow{ZY}=4\mathbf{a}\) and \(\overrightarrow{YX}=3\mathbf{b}\). The point P is such that WYP is a straight line and Y is the midpoint of WP. Show that WZ is parallel to XP.

• \(\overrightarrow{WZ}=\overrightarrow{WX}+\overrightarrow{XY}+\overrightarrow{YZ}=4\mathbf{a}-3\mathbf{b}\)
• \(\overrightarrow{WY}=\overrightarrow{WX}+\overrightarrow{XY}=8\mathbf{a}-3\mathbf{b},\quad \overrightarrow{YP}=\overrightarrow{WY}=8\mathbf{a}-3\mathbf{b}\)
• \(\overrightarrow{XP}=\overrightarrow{XY}+\overrightarrow{YP}=8\mathbf{a}-6\mathbf{b}\)
• \(\overrightarrow{XP}=2\overrightarrow{WZ}\), so WZ is parallel to XP.