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Solving geometric problems

Solving geometric problems

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Solving geometric problems

​​In a nutshell

You can solve geometric problems involving vectors in 3D with the same theory as you have previously used for 2D vectors.



Geometric problems

Finding position vectors, parallel and scalar multiples of vectors and the magnitude of a vector can all help solve geometric problems with 3D vectors. Unlike vectors in 2D, it is not easy to accurately plot the coordinates and their associated vectors by hand, however a rough sketch can help visualise the problem given.


Example 1

Three points form a triangle with the following coordinates: A(−1,6,7)A (-1,6,7), B(1,9,5)B (1,9,5) and C(−1,12,7)C(-1,12,7).

a) Show that the triangle is isosceles.

b) Find the area of the triangle.

c) Find the coordinates of point DD, so that ABCDABCD forms a rhombus.


a) An isosceles triangle has two sides of equal length. Find the magnitude of the vectors which form the sides of the triangle AB→\overrightarrow{AB}, AC→\overrightarrow{AC} and BC→\overrightarrow{BC}.


​AB→=OB→−OA→=(195)−(−167)=(23−2)\begin{aligned}\overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\&= \begin{pmatrix}1 \\9\\5 \end{pmatrix} - \begin{pmatrix}-1 \\6\\7 \end{pmatrix}\\&=\begin{pmatrix}2 \\3\\-2 \end{pmatrix}\end{aligned}​​
​AC→=OC→−OA→=(−1127)−(−167)=(060)\begin{aligned}\overrightarrow{AC} &= \overrightarrow{OC} - \overrightarrow{OA} \\&= \begin{pmatrix}-1 \\12\\7 \end{pmatrix} - \begin{pmatrix}-1 \\6\\7 \end{pmatrix}\\&=\begin{pmatrix}0 \\6\\0 \end{pmatrix}\end{aligned}​​
​BC→=OC→−OB→=(−1127)−(195)=(−232)\begin{aligned}\overrightarrow{BC} &= \overrightarrow{OC} - \overrightarrow{OB} \\&= \begin{pmatrix}-1 \\12\\7 \end{pmatrix} - \begin{pmatrix}1 \\9\\5 \end{pmatrix}\\&=\begin{pmatrix}-2 \\3\\2 \end{pmatrix}\end{aligned}​​
​∣AB→∣=22+32+(−2)2=17|\overrightarrow{AB}| = \sqrt{2^2+3^2+ (-2)^2} = \sqrt{17}​​
​∣AC→∣=02+62+02=6|\overrightarrow{AC}| = \sqrt{0^2+6^2+ 0^2} = 6​​
​∣BC→∣=(−2)2+32+22=17|\overrightarrow{BC}| = \sqrt{(-2)^2+3^2+ 2^2} = \sqrt{17}​​


Since ∣AB→∣=∣BC→∣|\overrightarrow{AB}| =|\overrightarrow{BC}| , triangle ABCABC is isosceles.


b) Now that the equal sides are known, a sketch can help visualise the triangle to help find the area:

Maths; Vectors II; KS5 Year 13; Solving geometric problems


Find the perpendicular height using Pythagoras' theorem:

h=(17)2−32=22h = \sqrt{(\sqrt{17})^2 - 3^2} = 2\sqrt2​​


Calculate the area

A=12bh=12×6×22=62‾\begin {aligned}A &= \dfrac 1 2 bh \\ \\&= \dfrac 1 2 \times 6 \times 2 \sqrt2 \\\\& = \underline{6\sqrt 2}\end{aligned}​​


c) Add point DD to the sketch.

Maths; Vectors II; KS5 Year 13; Solving geometric problems

From the sketch, it can be seen that for ABCDABCD to form a rhombus, AD→=BC→\overrightarrow{AD} = \overrightarrow{BC} and AB→=DC→\overrightarrow{AB} = \overrightarrow{DC}. Use this to find the position vector of DD:

​OD→=OA→+AD→=OA→+BC→=(−167)+(−232)=(−399)\begin{aligned}\overrightarrow{OD} &= \overrightarrow{OA}+ \overrightarrow{AD} \\&= \overrightarrow{OA}+ \overrightarrow{BC} \\& = \begin{pmatrix}-1 \\ 6 \\7 \end{pmatrix} + \begin{pmatrix}-2 \\ 3 \\2 \end{pmatrix} \\&= \begin{pmatrix}-3 \\ 9 \\9 \end{pmatrix}\end{aligned}​​


Therefore, DD has coordinates (−3,9,9)‾\underline{ (-3,9,9)}.

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Learn with Basics

Length:
Position vectors and displacement vectors

Unit 1

Position vectors and displacement vectors

Solving geometric problems

Unit 2

Solving geometric problems

Jump Ahead

Solving geometric problems

Unit 3

Solving geometric problems

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FAQs - Frequently Asked Questions

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