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

In triangle PQR, \(\overrightarrow{PQ}=3\mathbf{a}\) and \(\overrightarrow{PR}=2\mathbf{b}\). The point S lies on QR with QS:SR = 1:2. The point T is such that \(\overrightarrow{QT}=\frac{1}{2}\overrightarrow{PR}\). Show that P, S and T lie on a straight line.

• \(\overrightarrow{QR}=\overrightarrow{PR}-\overrightarrow{PQ}=-3\mathbf{a}+2\mathbf{b}\)
• \(\overrightarrow{PS}=\overrightarrow{PQ}+\frac{1}{3}\overrightarrow{QR}=2\mathbf{a}+\frac{2}{3}\mathbf{b}=\frac{2}{3}(3\mathbf{a}+\mathbf{b})\)
• \(\overrightarrow{PT}=\overrightarrow{PQ}+\overrightarrow{QT}=\overrightarrow{PQ}+\frac{1}{2}\overrightarrow{PR}=3\mathbf{a}+\mathbf{b}\)
• \(\overrightarrow{PT}=\frac{3}{2}\overrightarrow{PS}\). These are parallel and share P, so P, S and T are collinear.

In triangle OAB, \(\overrightarrow{OA}=8\mathbf{a}\) and \(\overrightarrow{OB}=5\mathbf{b}\). X is the midpoint of OA and Y is the midpoint of AB. Show that XY is parallel to OB.

• \(\overrightarrow{XA}=\frac{1}{2}\overrightarrow{OA}=4\mathbf{a}\)
• \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-8\mathbf{a}+5\mathbf{b},\quad \overrightarrow{AY}=\frac{1}{2}\overrightarrow{AB}=-4\mathbf{a}+\frac{5}{2}\mathbf{b}\)
• \(\overrightarrow{XY}=\overrightarrow{XA}+\overrightarrow{AY}=\frac{5}{2}\mathbf{b}\)
• \(\overrightarrow{XY}=\frac{1}{2}\overrightarrow{OB}\), so XY is parallel to OB

Show that the points P(−2, 3, 1), Q(1, 5, −3) and R(4, 7, −7) are collinear.

• \(\overrightarrow{PQ}=\begin{pmatrix}1\\5\\-3\end{pmatrix}-\begin{pmatrix}-2\\3\\1\end{pmatrix}=\begin{pmatrix}3\\2\\-4\end{pmatrix}\)
• \(\overrightarrow{PR}=\begin{pmatrix}4\\7\\-7\end{pmatrix}-\begin{pmatrix}-2\\3\\1\end{pmatrix}=\begin{pmatrix}6\\4\\-8\end{pmatrix}\)
• \(\overrightarrow{PR}=2\overrightarrow{PQ}\). The vectors are parallel and share P, so the points are collinear.

A roof truss ABC is modelled as a triangle, \(\overrightarrow{AB}=4\mathbf{a}\) and \(\overrightarrow{AC}=3\mathbf{b}\). The point D lies on BC with BD:DC = 3:1. The point E is such that \(\overrightarrow{BE}=3\overrightarrow{AC}\). Show that A, D and E lie on a straight line.

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

A tabletop KLMN is a trapezium (lengths in decimetres), \(\overrightarrow{KL}=6\mathbf{a}\), \(\overrightarrow{NM}=3\mathbf{a}\) and \(\overrightarrow{KN}=\mathbf{a}+5\mathbf{b}\). H is the midpoint of KL. Show that NH is parallel to ML.

• \(\overrightarrow{LM}=\overrightarrow{LK}+\overrightarrow{KN}+\overrightarrow{NM}=-2\mathbf{a}+5\mathbf{b}\)
• \(\overrightarrow{KH}=\frac{1}{2}\overrightarrow{KL}=3\mathbf{a}\)
• \(\overrightarrow{NH}=\overrightarrow{NK}+\overrightarrow{KH}=2\mathbf{a}-5\mathbf{b}\)
• \(\overrightarrow{NH}=-\overrightarrow{LM}=\overrightarrow{ML}\), so NH is parallel to ML.

A submarine is detected at three points D(1, −4, −6), E(3, −1, −2) and F(7, 5, 6) (coordinates in metres). Show that the submarine moved in a straight line.

• \(\overrightarrow{DE}=\begin{pmatrix}3\\-1\\-2\end{pmatrix}-\begin{pmatrix}1\\-4\\-6\end{pmatrix}=\begin{pmatrix}2\\3\\4\end{pmatrix}\)
• \(\overrightarrow{DF}=\begin{pmatrix}7\\5\\6\end{pmatrix}-\begin{pmatrix}1\\-4\\-6\end{pmatrix}=\begin{pmatrix}6\\9\\12\end{pmatrix}\)
• \(\overrightarrow{DF}=3\overrightarrow{DE}\). The vectors are parallel and share D, so the points are collinear.

In triangle ABC, \(\overrightarrow{AB}=3\mathbf{a}\) and \(\overrightarrow{AC}=4\mathbf{b}\). The point D lies on BC with BD:DC = 1:2. 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.

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

WXYZ is a quadrilateral, \(\overrightarrow{WX}=10\mathbf{a}\), \(\overrightarrow{ZY}=5\mathbf{a}\) and \(\overrightarrow{YX}=4\mathbf{b}\). 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.

• \(\overrightarrow{WZ}=\overrightarrow{WX}+\overrightarrow{XY}+\overrightarrow{YZ}=5\mathbf{a}-4\mathbf{b}\)
• \(\overrightarrow{WY}=\overrightarrow{WX}+\overrightarrow{XY}=10\mathbf{a}-4\mathbf{b},\quad \overrightarrow{YP}=\overrightarrow{WY}=10\mathbf{a}-4\mathbf{b}\)
• \(\overrightarrow{XP}=\overrightarrow{XY}+\overrightarrow{YP}=10\mathbf{a}-8\mathbf{b}\)
• \(\overrightarrow{XP}=2\overrightarrow{WZ}\), so WZ is parallel to XP.