9Bpositionvectors


 * Magnitude and Direction of Vectors**

**Magnitude** of a vector can be found using Pythagoras.
 * {compare with finding the modulus (absolute value) of a complex number. See 3D Polar Form}

math \\ . \qquad \underline{v}=a\underline{i}+b\underline{j} \qquad. \\ . \\ . \qquad v^2=|\underline{v}|^2=a^2+b^2 \qquad. \\ . \\ . \qquad v=\sqrt{a^2+b^2} math

For a 3-Dimensional vector math \\ . \qquad \underline{v}=a\underline{i}+b\underline{j}+c\underline{k} \qquad .\\. \\ . \qquad v=\sqrt{a^2+b^2+c^2} math

**Direction** of a 2-Dimensional vector is given as the angle measured anticlockwise from the __**positive x-axis**__.
 * {compare with finding the argument of a complex number. See 3D Polar Form}

math \\ \text{For } \;\underline{v}=a\underline{i}+b\underline{j} \qquad. \\ . \\ . \qquad \theta=\text{ Tan}^{-1} \big(\frac{b}{a} \big) \qquad \lbrace \textit{adjust for the appropriate quadrant} \rbrace \qquad. math

**NB:** There is no strictly enforced **Principal Domain** for direction of vectors.

For a 3-D vector, 2 angles are needed -- an angle in the x-y plane measured anticlockwise from the __**positive x-axis**__, then the angle measured vertically from that place in the plane up into the z-direction. {Not part of the course}

Example

Find the magnitude and direction for **__v__** = 6**__i__** + 3**__j__**

math \\ . \qquad |\underline{v}|=\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt{5}\qquad. \\ . \\ . \qquad \theta=\text{ Tan}^{-1} \big(\frac{3}{6} \big)=26.6^{\circ} \qquad. math

On your calculator, both of these operations can be done in one step using the **toPol** command from the ACTION menu, VECTOR submenu

Rectangular Form Rectangular Form is the name for **__i__** and __**j**__ notation.

Given a magnitude and direction, to get a vector in the form **__v__** = x**__i__** + y**__j__**

Use the rules: math \\ .\qquad x = r \, \cos \big( \theta \big) \qquad. \\ . \\ . \qquad y = r \, \sin \big( \theta \big) \qquad. math

Rectangular Form on the Calculator Again, use the single line matrix form in **2D** tab, **CALC** section. The angle symbol is available in the **mth** tab, **OPTN** section. math . \qquad \big[ 5 \; \angle 160 \big] math Just enter the matrix and hit EXE.

With the calculator in **Standard** mode, you will get: math . \qquad \big[ 5 \centerdot \cos (160) \; 5 \centerdot \sin (160) \big] \qquad. math

Change to **Decimal** mode and hit EXE again, and you will get: math . \qquad \big[ -4.70 \; 1.71 \big] \qquad. math

This is equivalent to the vector: **__v__** = –4.70**__i__** + 1.71**__j__**

NB: The **toRect** function will do the same thing, but it is not needed.

True Bearings

**True Bearings** are measured __clockwise__ from North. We treat the positive y-axis as North so measure True Bearings clockwise from the positive y-axis.

This means math \\ \text{If } \theta \leqslant 90^{\circ} \quad \text{True Bearing } = 90^{\circ}-\theta \qquad. \\ .\\ \text{If } \theta > 90^{\circ} \quad \text{True Bearing } = 360^{\circ}- \big( \theta-90^{\circ} \big) \qquad. math

Example 1

A hiker leaves camp and walks 50km due East and 18km due South. Express as a vector and hence find the final distance and True Bearing from the camp.

math \\ . \qquad \underline{v}=50\underline{i}-18\underline{j} \qquad. \\ . \\ . \qquad |\underline{v}|=\sqrt{50^2+(-18)^2}=\sqrt{2824}=53.1 \text{ km} \qquad. \\ . \\ . \qquad \theta=\text{ Tan}^{-1} \big(\frac{-18}{50} \big) =-20^{\circ} \qquad \lbrace \textit{4th Quadrant} \rbrace \qquad. \\ . \\ \textit{so } \\. \\ . \qquad \text{Bearing } = 90-(-20)=110^{\circ} \text{ True} \qquad. math

Example 2

Consider the vector, **__u__**, with a magnitude of 30 and a True Bearing of 310 o. Express **__u__** in the from **__u__** = x**__i__** + y**__j__**.

Given the True Bearing of 310 o , this is 360 – 310 = 50 o anticlockwise from the positive y-axis {in 2nd Quadrant} math \text{so } \theta=90+50=140^{\circ} \qquad. math

Use the rules: math \\ . \qquad x=|\underline{u}|\cos(\theta) \qquad. \\ . \qquad y=|\underline{u}|\sin(\theta) math {same as the equivalent rules for complex numbers}

math \\ . \qquad x=30\cos(140^{\circ}) = -22.98 \qquad. \\ . \\ . \qquad y=30\sin(140^{\circ}) = 19.28 \\. \\ . \qquad \underline{u}=-22.98\underline{i}+19.28\underline{j} \qquad. math

Vectors on the Classpad

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