Unit 3

The magnitude, or absolute value, of a vector is its length.
A unit vector is a vector that is 1 unit long. Vectors i and j are unit vectors in the x- and y-directions, respectively.
A scalar is a quantity having only magnitude, not direction.
A position vector, starts at the origin and ends at a point (x, y)
A displacement vector is the difference between an object’s initial and final position
The sum, called the resultant vector, is the vector that goes from head to tail, starting from the beginning of the first vector to the end of the second

Example Problem- Component Form of Vectors

For Problems 9, resolve the vector into horizontal and vertical components

Dot Product

The dot product, or scalar product, is used to find the measure of the angle between two vectors or determining the effect of a force vector on another vector.

The equation of dot product is a・b = |a|*|b|cos (θ) where θ is the angle between them.

a・b is a number. The way that you find it is first by multiplying the coefficients of the two vectors i's together, the coefficients of the two vectors j's together, and the coefficients of the two vectors k's together. These three values are then added together to form the final number.

For example with vectors a = <1i + 3j + 2k> and b = <4i + 2j + 5k>

a・b = (1*4) + (3*2) + (2*5) = 20

Practice Problem #1: Vector Addition & Dot Product

#19- Vectors.mp4

Refer to this video for an explanation to Practice Problem #1.

Voiced by Quinn Luong

Cross Product

Cross Product vs. Dot Product

Computation of Cross Products by Means of Determinants

You form the determinant by writing the three unit vectors (i, j, k) along the top row, the coefficients of the first vector in the middle row, and the coefficients of the second vector in the bottom row.

To expand this determinant to find the first value of i, mentally cross out its row and its column, and multiply by the second-order determinant that remains. You do the same for j and k. The signs of the expanded determinant alternate, so the second term (j) has a - (negative) sign and the third term (k) has a + (positive) sign

Why, you ask? It's sadly "convention"

Example Problem- Solving Cross Product by Determinants

Example Problem explanation:

First, we are given the cross product and only one of the vectors. We can simply set up a determinant and solve the i, j, k terms. Solving for the i term, mentally erase that column and multiply with the criss-cross technique.

We get (4z-6)i - (3z-6x)j + (3-4x)k. Notice how they only ask to solve for x and z because j is equal to 1.

Using simple algebra, we set our equations equal to the correspondent coefficients of i, j, k of the cross product vector. We get z=2 and x=5.

Practice Problem #2: Cross Product Real-World Application

Equations for Planes in Space

Given 3 points on the plane one can find the equation of said plane:

Find the displacement vectors between the three points. This is done by subtracting the x, y, and z values of one point from another given point. Typically you find the vector from the first to second point and the vector between the first and third point for simplicity.

2. Next, your goal is to find the normal vector to the plane because the coefficients in front of the vector components are identical to the coefficients in the equation of the plane. This is done by taking the cross product of the two vectors.

3. Now that you have the normal vector, you can substitute the i for x, j for y, and k for z. This gives you one side of the equation of the plane, but you still need to find the other side, represented by the variable D

In order to find D you plug in one of the points into the equation and simplify. This should leave you with a number with no variables.

Finally, you put the two sides of the equation together and you have successfully found the equation of a plane given three points.

Example Problem- Making a Plane

Given these 3 points, find the Equation of a Plane:

P₁= (-5, 7, 8)

P₂= (-3, 2, 7)

P₃= (3, 6, 5)

Direction Cosines

Direction angles are the angles between a position vector and the 3 different axises. Alpha (α) is for the x-axis, Beta (β) for the y, and Gamma (γ) for the z-axis.

Direction cosines are simply, the cosines of the direction angles and are written simply as C₁ is the cosine of Alpha, C₂ is the cosine of Beta and C₃ is the cosine of Gamma.

One of the most important aspects of this is the use of the pythagorean property where C₁²+ C₂²+ C₃²=1. This is expanded on the idea that C₁i + C₂j + C₃k = 1 is the unit vector in the same direction because the definition of the unit vector is that its magnitude is one, which is always true when you find the magnitude of that vector because of the previous Pythagorean property

Example Problem- Direction Cosines

Practice 1: Find the direction angles and the direction cosines of v = 13i - 6j + 18k

Practice 1 explanation:

First, we must find the magnitude of the given vector because that is the hypotenuse of the triangle and allows use to use the Pythagorean theorem to find the missing direction angles.

A simple way to think about the direction cosines is that it is the coefficient of the vector that corresponds to the Alpha (i), Beta (j), or Gamma (k) over the magnitude of the vectors.

Then, to solve for the direction angles or just simply Alpha, Beta, and Gamma we take the inverse cosine of both sides of the equation to get an angle. This is the angle between the vector and the corresponding axis.

Example Problem #2- Direction Cosines

Practice 2:

If α=152° and β = 73°, find γ

Practice 2 explanation:

Our primary goal in this problem is to solve for the missing angle and the simplest way to do that is to plug in the angles into the equation.

cos²(α) + cos²(β) + cos² (γ) = 1

After we do that step, we must rearrange the equation in order to get cos² (γ), or C₃²by itself in order to solve for it. This leaves us with some number (0.367) and we take the inverse cosine of both the positive and negative number in order to get our possible angles. This is because there are two possible values of γ.

Vectors Equations for Lines in Space

Equations for lines in space are the most complex looking problems through all of the vector unit, but their basic concept is quite simple. In essence, the equation of a line in space is made up of two parts. The first part is a position vector to a point on the line.

The second part is a unit vector in the direction of the line. This makes logical sense that it forms a line because following the rules of addition of vectors you put them head to tail so the unit vector would go some distance from a point on the line in the direction of the line. This means that it essentially maps out the path of the line, and with the addition of the variable d it is possible to write an equation for every point on the line.

Practice 1:

Find the equation of the line from points (5, 1, -4) to (14, 21, 8)

Practice 1 explanation:

The first step in order to find the equation of a line in space is to get the vector between the two points. The goal of finding this vector on the line is to get the unit vector on the line.

The next step to finding the unit vector is to get the magnitude of the vector between the two points.

After that we can put the coefficients over the magnitude of the vector in order to find the unit vector, and simplify the fractions.

Now we have all of the information required to find the vector equation of a line in space, but we still need to put it together. The vector P₀is the position vector to the line in space, which is the same as the vector to one of the points. We are going to choose the first point (5, 1, -4).

The second part of the equation is the unit vector that you just found. You also put on the units d because this serves the purpose of the x-value in a normal equation, to specify where on the line it is.

These two parts of the equation are then added together in parenthesis, both with their respective units (i, j, k) in order to form the full equation.

Practice 2:

Find the point of intersection between the line in the previous problem and the plane: 2x + 4y + 2z = 20

Practice 2 Explanation:

Our first goal in solving this problem is to find the value of d. The way that we go about doing this is by plugging in the coefficients of i, j, and k into the equation of the plane. This yields an equation with one variable, d.

We then fully simplify this equation and are left with the value of d.

Plugging the value of d back into the original equation of the line were can then compute the values of the point of intersection. This is because, when simplified, this equation would yield the vector to the point of intersection so if we take the coefficients we can put them together and find the point.

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