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Vectors II

Solving geometric problems

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Further kinematics

Vector methods with projectiles

Variable acceleration in one dimension

Differentiating vectors

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Functions of time: Displacement, velocity, acceleration

Using differentiation to find velocity and acceleration

Maxima and minima problems

Using integration to find velocity and displacement

Constant acceleration formulae

Constant acceleration

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Constant acceleration: Deriving formulae

Constant acceleration formulae: SUVAT

Vertical motion under gravity

Modelling in mechanics

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Quantities and units in mechanics

Working with vectors

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The normal distribution: Finding probabilities

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Vectors II

3D coordinates

Vectors in 3D

Solving geometric problems

Applications to mechanics: 3D vectors

Numerical methods

Locating roots

Iteration

Staircase and cobweb diagrams

Newton-Raphson method

Numerical methods: Applications to modelling

Integration II

Integrating standard functions

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Reverse chain rule

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Finding areas using integration

The trapezium rule

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Differentiation II

Differentiating sin x and cos x

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Differentiating trigonometric functions

Parametric differentiation

Implicit differentiation

Second derivatives: Concave and convex functions

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Trigonometry and modelling

Addition formulae

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Simplifying a cos x + b sin x

Proving trigonometric identities

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Further trigonometric identities

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Solving trigonometric equations

Small angle approximations

Sequences and series

Sequences revision

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Binomial expansion II

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Functions

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The modulus function

y = |f(x)| and y = f(|x|)

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Modulus inequalities

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Proof II

Proof by contradiction

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Vectors I

Vectors

Representing vectors

Magnitude and direction of a vector

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Solving geometric problems

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Integration I

Integrating x^n

Indefinite integrals

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Differentiation I

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Proof I

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Summary

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: $A (-1,6,7)$ , $B (1,9,5)$ and $C(-1,12,7)$ .

a) Show that the triangle is isosceles.

b) Find the area of the triangle.

c) Find the coordinates of point $D$ , so that $ABCD$ 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 $\overrightarrow{AB}$ , $\overrightarrow{AC}$ and $\overrightarrow{BC}$ .

$\begin{aligned}\overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\&= \begin{pmatrix}1 \\9\\5 \end{pmatrix} - \begin{pmatrix}-1 \\6\\7 \end{pmatrix}\\&=\begin{pmatrix}2 \\3\\-2 \end{pmatrix}\end{aligned}$	$\begin{aligned}\overrightarrow{AC} &= \overrightarrow{OC} - \overrightarrow{OA} \\&= \begin{pmatrix}-1 \\12\\7 \end{pmatrix} - \begin{pmatrix}-1 \\6\\7 \end{pmatrix}\\&=\begin{pmatrix}0 \\6\\0 \end{pmatrix}\end{aligned}$	$\begin{aligned}\overrightarrow{BC} &= \overrightarrow{OC} - \overrightarrow{OB} \\&= \begin{pmatrix}-1 \\12\\7 \end{pmatrix} - \begin{pmatrix}1 \\9\\5 \end{pmatrix}\\&=\begin{pmatrix}-2 \\3\\2 \end{pmatrix}\end{aligned}$
$\|\overrightarrow{AB}\| = \sqrt{2^2+3^2+ (-2)^2} = \sqrt{17}$	$\|\overrightarrow{AC}\| = \sqrt{0^2+6^2+ 0^2} = 6$	$\|\overrightarrow{BC}\| = \sqrt{(-2)^2+3^2+ 2^2} = \sqrt{17}$

Since $|\overrightarrow{AB}| =|\overrightarrow{BC}|$ , triangle $ABC$ 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:

$h = \sqrt{(\sqrt{17})^2 - 3^2} = 2\sqrt2$ 

Calculate the area

$\begin {aligned}A &= \dfrac 1 2 bh \\ \\&= \dfrac 1 2 \times 6 \times 2 \sqrt2 \\\\& = \underline{6\sqrt 2}\end{aligned}$ 

c) Add point $D$ to the sketch.

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

$\begin{aligned}\overrightarrow{OD} &= \overrightarrow{OA}+ \overrightarrow{AD} \\&= \overrightarrow{OA}+ \overrightarrow{BC} \\& = \begin{pmatrix}-1 \\ 6 \\7 \end{pmatrix} + \begin{pmatrix}-2 \\ 3 \\2 \end{pmatrix} \\&= \begin{pmatrix}-3 \\ 9 \\9 \end{pmatrix}\end{aligned}$

Therefore, $D$ has coordinates $\underline{ (-3,9,9)}$ .

Learn with Basics

Length:

Unit 1

Position vectors and displacement vectors

Unit 2