10.8 Vectors  (Page 4/22)

We also have v ₂ with initial point $\left(0,0\right)$ and terminal point $\left(0,3\right).$

Therefore, the position vector is

Using the Pythagorean Theorem, the magnitude of v ₁ is 2, and the magnitude of v ₂ is 3. To find the magnitude of v , use the formula with the position vector.

The magnitude of v is $\sqrt{13}.$ To find the direction, we use the tangent function $\tan \theta =\frac{y}{x}.$

Thus, the magnitude of $v$ is $\sqrt{13}$ and the direction is ${56.3}^{\circ }$ off the horizontal.

Finding the components of the vector

Find the components of the vector $v$ with initial point $\left(3,2\right)$ and terminal point $\left(7,4\right).$

First find the standard position.

$$\begin{array}{l}v=⟨7-3,4-2⟩\hfill \\ =⟨4,2⟩\hfill \end{array}$$

See the illustration in [link] .

The horizontal component is $v_1$ $=⟨4,0⟩$ and the vertical component is $v_2$ $=⟨0,2⟩.$

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Finding the unit vector in the direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector    . We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as $i$ $=⟨1,0⟩$ and is directed along the positive horizontal axis. The vertical unit vector is written as $j$ $=⟨0,1⟩$ and is directed along the positive vertical axis. See [link] .

Finding the unit vector in the direction of v

Find a unit vector in the same direction as $v$ $=⟨\mathrm{-5},12⟩.$

First, we will find the magnitude.

$$\begin{array}{l}\left|v\right|=\sqrt{{(-5)}^2+{(12)}^2}\hfill \\ =\sqrt{25+144}\hfill \\ =\sqrt{169}\hfill \\ =13\hfill \end{array}$$

Then we divide each component by $|v|,$ which gives a unit vector in the same direction as v :

$$\frac{v}{\left|v\right|}=-\frac{5}{13}i+\frac{12}{13}j$$

or, in component form

$$\frac{v}{\left|v\right|}=⟨-\frac{5}{13},\frac{12}{13}⟩$$

See [link] .

Verify that the magnitude of the unit vector equals 1. The magnitude of $-\frac{5}{13}i+\frac{12}{13}j$ is given as

$$\begin{array}{l}\sqrt{{\left(-\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2}=\sqrt{\frac{25}{169}+\frac{144}{169}}\hfill \\ =\sqrt{\frac{169}{169}}=1\hfill \end{array}$$

The vector u $=\frac{5}{13}$ i $+\frac{12}{13}$ j is the unit vector in the same direction as v $=⟨-5,12⟩.$

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Performing operations with vectors in terms of i And j

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j .

a general note label

Vectors in the rectangular plane

Given a vector $v$ with initial point $P=\left(x_1,y_1\right)$ and terminal point $Q=(x_2,y_2),$ v is written as

$v=\left(x_2-x_1\right)i+\left(y_1-y_2\right)j$

The position vector from $\left(0,0\right)$ to $\left(a,b\right),$ where $\left(x_2-x_1\right)=a$ and $\left(y_2-y_1\right)=b,$ is written as v = a i + b j . This vector sum is called a linear combination of the vectors i and j .

The magnitude of v = a i + b j is given as $\left|v\right|=\sqrt{a^2+b^2}.$ See [link] .