9Ecomponents

When a vector, **__v__**, is seperated into two parts (addition in reverse), they are called the **components** of the vector.
 * Components of a Vector**

If the components of the vector are perpendicular, they are called **resolutes** .

If **__a__** and **__b__** are perpendicular, and **__v__** = **__a__** + **__b__** then using trigonometry, we get: math \\ . \qquad a=v\cos \, ( \theta ) \qquad. \\ . \qquad b=v\sin \, ( \theta ) math

The most commonly used resolutes ( perpendicular components) of a vector are the horizontal and vertical components (multiples of **__i__** and **__j__**)

Example 1 The horizontal and vertical components of **__v__** = –4**__i__** + 3**__j__** are ... ... ... **__x__** = –4**__i__** ... and ... ... ... **__y__** = 3**__j__**

Example 2 A vector of magnitude 5 makes an angle of 30 o with the positive x-axis. Find the horizontal and vertical components and thus express in the form, **__v__** = x**__i__** + y**__j__**

math \\ . \qquad x=5\cos 30^{\circ} = \dfrac{5\sqrt{3}}{2} \qquad. \\ . \\ . \qquad y=5\sin 30^{\circ} = \dfrac{5}{2} \qquad. \\ . \\ \text{so} \\. \\ \underline{v}=\frac{1}{2} \big( 5\sqrt{3} \, \underline{i}+5 \, \underline{j} \big) \qquad. math

Note: Adding two vectors is easier if they are converted into the vertical and horizontal components first. Example 3 Vector **__v__** has magnitude 4 in the direction N30 o E and vector **__w__** has magnitude 5 in the direction S60 o E

a) Find **__v__** + __**w**__ **For v:** math \\ \theta=60^{\circ} \quad \lbrace \textit{1st Quadrant} \rbrace \qquad . \\ . \\ . \qquad x=4\cos 60^{\circ}=\frac{4}{2}=2 \\ . \\ . \qquad y=4\sin 60^{\circ}=\frac{4}{2}\sqrt{3}=2\sqrt{3} \qquad . \\ . \\ \text{so} \\. \\ \underline{v}=2\underline{i}+2\sqrt{3}\, \underline{j} \qquad . math

**For w:** math \\ \theta=-30^{\circ} \quad \lbrace \textit{4th Quadrant} \rbrace \qquad. \\ . \\ . \qquad x=5\cos (-30^{\circ})=\frac{5}{2}\sqrt{3} \\. \\ . \qquad y=5\sin (-30^{\circ})=-\frac{5}{2} \\. \\ \text{so} \\. \\ \underline{w}=\frac{5}{2}\sqrt{3} \, \underline{i}-\frac{5}{2} \, \underline{j} \qquad. math

**Thus __v__ + __w__:** math \\ =\big( 2\underline{i}+2\sqrt{3}\, \underline{j} \big)+\big( \frac{5}{2}\sqrt{3} \, \underline{i}-\frac{5}{2} \, \underline{j} \big) \qquad. \\ . \\ =\big( 2+\frac{5}{2}\sqrt{3} \big) \underline{i} + \big( 2\sqrt{3}-\frac{5}{2} \big) \underline{j} \qquad. math

b) Round answers to 3 decimal places and use to find the magnitude and direction of **__v__** + **__w__** math \\ \underline{v}+\underline{w}=6.330\underline{i}+0.964\underline{j} \qquad .\\ . \\ . \qquad =6.403 \qquad . math
 * \underline{v}+\underline{w}|=\sqrt{6.330^2+0.964^2} \qquad . \\ . \\

Magnitude = 6.403

math \theta=\text{Tan}^{-1} \big( \frac{0.964}{6.330} \big) \qquad. \\ . \\ \theta = 8.66^{\circ} \qquad. math {anticlockwise from positive x-axis} BUT: Use the same notation as the question so change to Bearings.

Direction = N81.34 o E

Magnitude = 6.403 .