Vectors - the Components

Vectors - the Components - How it Works - Video

Vectors - Definitions - Part 1

Vectors - Definitions - Part 1:

The magnitude helps us find the distance of our vector.

We have 3 special vectors the 0 vector the i vector and j vector. The 0 vector is <0, 0>; the i is <1, 0>; the j is <0, 1>.

Vectors - Definitions - Part 2

Vectors - Definitions - Part 2:

Here we have more definitions.

We have added cosine and sine.

Using SOH CAH TOA and the fact that the i vector is the horizontal component and the j vector is the vertical component. We can substitute multiple times to get the formulas above.

Example 1

Example 1:

Here we the wind blowing in the direction of N30°W. The red line represents this. We have to be careful here and making sure that we start at the right spot otherwise the answer will be incorrect.

N is where we start, so we go in the positive y-axis direction for north, then we go 30° from the positive y-axis towards west. Now that we have drawn where the wind is blowing, we can separate it into the i and j components.

Remember v1 = ||a|| * cosθ and v2 = ||a|| * sinθ. In our example the mi/hr is our magnitude, so ||a|| = 15. Now we have to find our θ. The θ that we want starts from the positive x-axis and goes to the red line. Therefore θ = 90° + 30° => 120°.

So v1 = 15 * cos120° and v2 = 15 * sin120°.

Now we look back at our unit circle so find the values in order to get an exact answer so our final answer is v = -7.5i + [15sqrt(3)/2] j.

Example 2

Example 2:

This one we have several steps to find our final answer. Let's write down what we need: the magnitude of a, and the unit vector.

Let's find the magnitude of a.

||a|| = sqrt[ a12 + a22 ]

= sqrt[62 + (-8)2]

= sqrt[36 + 64]

= sqrt[100]

= 10


Substituted for a1 and a2.

Squared both numbers.

Found the sum.

Square rooted.

Now that we have found the magnitude of a. We need to find the unit vector so that vector is in direction form. Remember vectors gives direction and magnitude. When we find the unit vector, the magnitude is 1, so it only tells us the direction, hence direction form.

Let's find our unit vector so we are going to by 1 over the magnitude of that we found, 1/||a||.

unit vector = 1/||a|| * < a1 + a1 >

= [1/10] * <6, -8>

= <[1/10] * 6, [1/10] * (-8)>

<6/10, -8/10>

<3/5, -4/5>


Substituted for a1, a2, and the magnitude.

Distributed the scalar multiple.

Multiplied.

Simplified.

Now we have our unit vector <3/5, -4/5>. It is still going in the same direction as our original vector <6, -8>. And our magnitude of <3/5, -4/5> is 1.

We still need to find our vector b.

We want to go in the opposite direction with a magnitude of 7 so let's multiply our unit vector by the scalar multiple of -7.

b = -7 * u

= -7 * <3/5, -4/5>

= <-7 * (3/5), -7 * (-4/5)>

<-21/5, 28/5>


Substituted the unit vector.

Distributed the scalar multiple.

Multiplied.

So b = <-21/5, 28/5>.

Example 3

Example 3:

In this example we have to add vectors to find our resultant vector.

First we need to find our horizontal and vertical components of vectors MN and MO.

For MN N is where we start, so we go in the positive y-axis direction for north, then we go 20° from the positive y-axis towards east. For NO N is where we start, so we go in the positive y-axis direction for north, then we go 65° from the positive y-axis towards east.

Remember when we draw our angles, we start at the positive x-axis and go to the line. For the red this makes the angle 70° and for the blue line the angle is 25°. We use those to create our horizontal and vertical components.

MN = (6 * cos70° ) * i + (6 * sin70° ) * j

MO = (10 * cos25° ) * i + (10 * sin25° ) * j

Example 3:

Now that have our horizontal and vertical components for each vector. Let's combine them.

MN = (6 * cos70° ) * i + (6 * sin70° ) * j

MO = (10 * cos25° ) * i + (10 * sin25° ) * j

MN + MO = (6 * cos70° ) * i + (10 * cos25° ) * i + (6 * sin70° ) * j + (10 * sin25° ) * j

MN + MO = (6 * cos70° + 10 * cos25° ) * i + (6 * sin70° + 6 * sin70° ) * j

MN + MO = 11.1 * i + 9.9 * j

We need to find the magnitude of ||MS|| => sqrt[ a12 + a22 ] => sqrt[ 11.12 + 9.92 ] = 14.9.

Now we need to find the angle created by our magnitude of ||MS||. We can use tangent to find that,

tanθ = opposite/adjacent => 9.9/11.1

θ = tan-1(9.9/11.1)

θ = 41.6°

Now we need to find the angle that is created on the opposite side so we can use N in our answer. So we subtract 90° and 41.6° to get 48.4°.

So we have N48.4°E.

If you wanted to you use 41.6° in your answer by saying E41.6°N, but that depends on the specifics of the question or your teacher.

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