Lesson

Vectors are often expressed in the form of coordinates. In two dimensions, we could have a vector $(a,b)$(`a`,`b`). Its equivalent in the form of an arrow has the tail of the arrow at the origin $(0,0)$(0,0) and the arrow head at the point $(a,b)$(`a`,`b`).

It is useful to remember that an arrow vector can be translated anywhere in the space, keeping its direction and length fixed, and it is still considered to be the same vector. For example, if we want the tail of the vector $(a,b)$(`a`,`b`) to be at the point $(-1,6)$(−1,6), say, then the head is at the point $(a-1,b+6)$(`a`−1,`b`+6).

In geometrical problems and proofs, we can interpret line segments as vectors.

A line segment drawn between two points whose coordinates are given is equivalent to the vector formed by the difference between the vectors defined by the given points. The following diagram illustrates this idea.

In the illustration, the line segment AB is equivalent to the vector (6, -0.5) by translation. We obtain it by calculating $\overrightarrow{OB}-\overrightarrow{OA}=(7,3)-(1,3.5)$›‹`O``B`−›‹`O``A`=(7,3)−(1,3.5).

The coordinate form $(a,b)$(`a`,`b`) of a vector can be understood as the vector sum $a\mathbf{i}+b\mathbf{j}$`a`** i**+

To perform a vector addition or subtraction when the vectors are given in the form of magnitudes and directions (that is, as arrows) we should convert the vectors into coordinate form first. This may involve some trigonometry.

Two velocity vectors are separated by an angle of $45^\circ$45°. One has a magnitude of $100$100 m/s and the other, with the angle measured anticlockwise from the first, has a magnitude of $75$75 m/s. Find the resultant when these two vectors are added.

The coordinates of the vector with magnitude $75$75 m/s, are found by trigonometry to be $(75\cos45^\circ,75\sin45^\circ)$(75`c``o``s`45°,75`s``i``n`45°). That is, $\left(\frac{75}{\sqrt{2}},\frac{75}{\sqrt{2}}\right)$(75√2,75√2). Then, the sum of the two vectors is the vector $\left(100+\frac{75}{\sqrt{2}},0+\frac{75}{\sqrt{2}}\right)\approx\left(153,53\right)$(100+75√2,0+75√2)≈(153,53).

By Pythagoras, we see that this resultant vector has magnitude $\sqrt{153^2+53^2}\approx161.9$√1532+532≈161.9 m/s. Its angle from the $100$100 m/s vector is $\arcsin\left(\frac{53}{161.9}\right)=19.1^\circ$`a``r``c``s``i``n`(53161.9)=19.1°.

For many problems to do with vectors, it is sufficient to understand the basic ideas presented in other chapters about magnitude, components, addition and subtraction, and scalar multiplication.

Vectors $U$`U`, $V$`V` and $W$`W` form three sides of a triangle, as shown in the graph below.

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Find an expression for $W$

`W`in terms of $U$`U`and $V$`V`:Now find an expression for $V$

`V`in terms of $U$`U`and $W$`W`:

Consider vectors $V$`V` and $U$`U` of magnitude $4$4 and $5.8$5.8 units respectively, separated by an angle of $31^\circ$31°, as shown on the graph. Plot the resultant vector $W=U+V$`W`=`U`+`V` with its tail at the origin.

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