Vectors in geometry

Equal and opposite vectors, addition, midpoints, centroid, points dividing a segment, the regular hexagon. 12 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Equal and opposite vectors

\(ABCD\) is a parallelogram and \(O\) is the intersection of its diagonals.

a) Name all the vectors equal to \(\vec{AO}\) and to \(\vec{AB}\). b) Name all the vectors opposite to \(\vec{OB}\). c) Is \(\vec{AD} = \vec{BC}\)? Is \(\vec{AD} = \vec{CB}\)?

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The diagonals of a parallelogram bisect each other, so \(AO = OC\) and \(BO = OD\).

a) \(\vec{AO} = \vec{OC}\); \(\vec{AB} = \vec{DC}\).
b) Opposite vectors have the same length and opposite direction: \(\vec{BO}\) and \(\vec{OD}\).
c) \(\vec{AD} = \vec{BC}\) (same length, same direction). \(\vec{AD} \ne \vec{CB}\): they are opposite, \(\vec{CB} = -\vec{AD}\).

Problem 2 Midpoints of a triangle

In triangle \(ABC\), the points \(K\), \(L\), \(M\) are the midpoints of \(AB\), \(BC\) and \(CA\). Name all the vectors (with endpoints among \(A, B, C, K, L, M\)) equal to \(\vec{KL}\), to \(\vec{AM}\) and to \(\vec{LM}\).

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By the midline theorem \(KL \parallel AC\), \(KL = \frac12 AC\); \(LM \parallel AB\), \(LM = \frac12 AB\); \(KM \parallel BC\), \(KM = \frac12 BC\).

\(\vec{KL} = \vec{AM} = \vec{MC}\) (all point from the side \(A\) towards \(C\), half of \(AC\)).
\(\vec{AM} = \vec{KL} = \vec{MC}\).
\(\vec{LM} = \vec{BK} = \vec{KA}\).

Problem 3 Adding vectors

Simplify.

a) \(\vec{AB} + \vec{BC} + \vec{CD}\) b) \(\vec{PQ} + \vec{RP} + \vec{QS}\) c) \(\vec{MN} - \vec{KN} + \vec{KL}\) d) \(\vec{OA} - \vec{OB} + \vec{BC} - \vec{AC}\)

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Use \(\vec{XY} + \vec{YZ} = \vec{XZ}\) and \(-\vec{XY} = \vec{YX}\).

a) \(\vec{AD}\)
b) \(\vec{RP} + \vec{PQ} + \vec{QS} = \vec{RS}\)
c) \(\vec{MN} + \vec{NK} + \vec{KL} = \vec{ML}\)
d) \(\vec{OA} - \vec{OB} = \vec{BA}\), so we get \(\vec{BA} + \vec{BC} + \vec{CA}\). Since \(\vec{BC} + \vec{CA} = \vec{BA}\), the result is \(2\vec{BA}\).

Problem 4 Median of a triangle

In triangle \(ABC\), \(M\) is the midpoint of \(BC\). Express \(\vec{AM}\) in terms of \(\vec{AB}\) and \(\vec{AC}\).

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\(\vec{AM} = \vec{AB} + \vec{BM} = \vec{AB} + \frac12\vec{BC}\) and \(\vec{BC} = \vec{AC} - \vec{AB}\).

\[\vec{AM} = \vec{AB} + \tfrac12(\vec{AC} - \vec{AB}) = \tfrac12\vec{AB} + \tfrac12\vec{AC}.\]

Problem 5 Point dividing a side

In triangle \(ABC\), the point \(D\) lies on \(BC\) with \(|BD| = 3|DC|\). Express \(\vec{AD}\) in terms of \(\vec{AB}\) and \(\vec{AC}\).

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\(|BD| = \frac34|BC|\), so \(\vec{BD} = \frac34\vec{BC} = \frac34(\vec{AC} - \vec{AB})\).

\[\vec{AD} = \vec{AB} + \vec{BD} = \tfrac14\vec{AB} + \tfrac34\vec{AC}.\]

Problem 6 Centroid

\(G\) is the centroid of triangle \(ABC\) and \(P\) is any point. Show that \(\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}\) and that \(\vec{PA} + \vec{PB} + \vec{PC} = 3\vec{PG}\).

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Let \(M\) be the midpoint of \(BC\). Then \(\vec{GB} + \vec{GC} = 2\vec{GM}\). The centroid divides the median \(AM\) in the ratio \(2:1\), so \(\vec{GA} = -2\vec{GM}\).

Therefore \(\vec{GA} + \vec{GB} + \vec{GC} = -2\vec{GM} + 2\vec{GM} = \vec{0}\).

For any \(P\): \(\vec{PA} + \vec{PB} + \vec{PC} = 3\vec{PG} + (\vec{GA} + \vec{GB} + \vec{GC}) = 3\vec{PG}\).

Problem 7 Points on a line

The points \(A\), \(C\), \(M\), \(B\) lie on a line in this order, \(|AB| = 4|AC|\) and \(M\) is the midpoint of \(CB\). Express \(\vec{MA}\) in terms of \(\vec{MC}\) and \(\vec{MB}\).

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Take \(|AC| = 1\), so \(|AB| = 4\) and \(|CB| = 3\). Then \(|MC| = |MB| = 1.5\) and \(|MA| = 2.5\).

\(\vec{MA}\) points the same way as \(\vec{MC}\) (towards \(A\)), so \(\vec{MA} = \frac{2.5}{1.5}\vec{MC} = \frac53\vec{MC}\). Since \(\vec{MB} = -\vec{MC}\), we can also write \(\vec{MA} = -\frac53\vec{MB}\).

A form with both vectors: \(\vec{MA} = \frac43\vec{MC} - \frac13\vec{MB}\) (check: \(\frac43\vec{MC} + \frac13\vec{MC} = \frac53\vec{MC}\)).

Problem 8 Regular hexagon

\(ABCDEF\) is a regular hexagon with centre \(O\), \(\vec{AB} = \vec{a}\) and \(\vec{BC} = \vec{b}\). Express in terms of \(\vec{a}\) and \(\vec{b}\):

a) \(\vec{CD}\) b) \(\vec{AO}\) c) \(\vec{AD}\) d) \(\vec{DE}\) e) \(\vec{AC}\) f) \(\vec{FC}\)

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In a regular hexagon the six triangles \(OAB, OBC, \dots\) are equilateral, so \(\vec{OC} = \vec{AB} = \vec{a}\), \(\vec{AF} = \vec{CD}\) and \(\vec{BC} = \vec{AO} = \vec{OD} = \vec{b}\).

a) \(\vec{CD} = \vec{BO} = \vec{BA} + \vec{AO} = -\vec{a} + \vec{b}\), so \(\vec{CD} = \vec{b} - \vec{a}\).
b) \(\vec{AO} = \vec{b}\).
c) \(\vec{AD} = 2\vec{AO} = 2\vec{b}\).
d) \(\vec{DE} = -\vec{AB} = -\vec{a}\).
e) \(\vec{AC} = \vec{a} + \vec{b}\).
f) \(\vec{FC} = 2\vec{AB} = 2\vec{a}\).

Problem 9 Midline with vectors

a) In triangle \(ABC\), \(D\) and \(E\) are the midpoints of \(AB\) and \(AC\). Show with vectors that \(DE \parallel BC\) and \(|DE| = \frac12|BC|\).

b) Show with vectors that the diagonals of a parallelogram bisect each other.

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a) \(\vec{DE} = \vec{DA} + \vec{AE} = \frac12\vec{BA} + \frac12\vec{AC} = \frac12(\vec{BA} + \vec{AC}) = \frac12\vec{BC}\). So \(\vec{DE}\) is a multiple of \(\vec{BC}\): the lines are parallel and \(|DE| = \frac12|BC|\).

b) Let \(ABCD\) be a parallelogram, \(\vec{AB} = \vec{u}\), \(\vec{AD} = \vec{v}\). The midpoint \(M\) of \(AC\) satisfies \(\vec{AM} = \frac12(\vec{u} + \vec{v})\). The midpoint \(N\) of \(BD\): \(\vec{AN} = \vec{AB} + \frac12\vec{BD} = \vec{u} + \frac12(\vec{v} - \vec{u}) = \frac12(\vec{u} + \vec{v})\). So \(M = N\).

Problem 10 Sum of medians

In triangle \(ABC\), \(D\), \(E\), \(F\) are the midpoints of \(BC\), \(CA\), \(AB\). Find \(\vec{AD} + \vec{BE} + \vec{CF}\).

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\(\vec{AD} = \frac12(\vec{AB} + \vec{AC})\), \(\vec{BE} = \frac12(\vec{BA} + \vec{BC})\), \(\vec{CF} = \frac12(\vec{CA} + \vec{CB})\).

In the sum, \(\vec{AB} + \vec{BA} = \vec{0}\), \(\vec{AC} + \vec{CA} = \vec{0}\), \(\vec{BC} + \vec{CB} = \vec{0}\). So \(\vec{AD} + \vec{BE} + \vec{CF} = \vec{0}\).

Problem 11 Ratios in a parallelogram

\(OPQR\) is a parallelogram with \(\vec{OP} = \vec{p}\) and \(\vec{OR} = \vec{r}\). \(S\) is the point on \(OP\) with \(OS : SP = 1 : 2\) and \(T\) is the midpoint of \(PQ\).

a) Express \(\vec{OS}\), \(\vec{OT}\) and \(\vec{ST}\) in terms of \(\vec{p}\) and \(\vec{r}\). b) \(U\) is on \(OQ\) with \(\vec{OU} = \frac34\vec{OQ}\). Express \(\vec{SU}\).

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a) \(\vec{OS} = \frac13\vec{p}\). \(\vec{PQ} = \vec{OR} = \vec{r}\), so \(\vec{OT} = \vec{p} + \frac12\vec{r}\). \(\vec{ST} = \vec{OT} - \vec{OS} = \frac23\vec{p} + \frac12\vec{r}\).

b) \(\vec{OQ} = \vec{p} + \vec{r}\), \(\vec{OU} = \frac34\vec{p} + \frac34\vec{r}\), \(\vec{SU} = \vec{OU} - \vec{OS} = \frac{5}{12}\vec{p} + \frac34\vec{r}\).

Problem 12 Intersection of lines

\(OAB\) is a triangle with \(\vec{OA} = \vec{a}\), \(\vec{OB} = \vec{b}\). \(M\) is the midpoint of \(OA\) and \(N\) is on \(OB\) with \(ON : NB = 2 : 1\). The lines \(AN\) and \(BM\) meet at \(X\). Find \(\vec{OX}\) in terms of \(\vec{a}\) and \(\vec{b}\).

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\(X\) on \(AN\): \(\vec{OX} = \vec{a} + s\left(\frac23\vec{b} - \vec{a}\right) = (1 - s)\vec{a} + \frac{2s}{3}\vec{b}\).

\(X\) on \(BM\): \(\vec{OX} = \vec{b} + t\left(\frac12\vec{a} - \vec{b}\right) = \frac{t}{2}\vec{a} + (1 - t)\vec{b}\).

\(\vec{a}\) and \(\vec{b}\) are not parallel, so the coefficients are equal: \(1 - s = \frac{t}{2}\) and \(\frac{2s}{3} = 1 - t\). From the second \(t = 1 - \frac{2s}{3}\); then \(1 - s = \frac12 - \frac{s}{3}\), so \(s = \frac34\) and \(t = \frac12\).

\(\vec{OX} = \frac14\vec{a} + \frac12\vec{b}\).