Algebraic Vectors and Parametric Equations - Pre-Calculus

First, you must find the length of the vector. This is given by the equation:

Then, dividing the vector by its length gives the unit vector in the same direction.

To get the unit vector that is in the same direction as the original vector , we divide the vector by the magnitude of the vector.

For , the magnitude is:

.

This means the unit vector in the same direction of is,

.

In order to find the unit vector u of a given vector v, we follow the formula

Let

The magnitude of v follows the formula

.

For this vector in the problem

.

Following the unit vector formula and substituting for the vector and magnitude

.

As such,

.

In order to find the unit vector u of a given vector v, we follow the formula

Let

The magnitude of v follows the formula

For this vector in the problem

Following the unit vector formula and substituting for the vector and magnitude

As such,

First, draw the vector = (3, 4); this is represented in red below. Then, plot the point P1 (1, 4), and draw a line (represented in blue) through it that is parallel to the vector .

We must find the equation of line . For any point P2 (x, y) on , . Since is on line and is parallel to , for some value of t. By substitution, we have . Therefore, the equation is a vector equation describing all of the points (x, y) on line parallel to through P1 (1, 4).

This is true. The independent variable in this equation is called a parameter.

A line through a point (x1,y1) that is parallel to the vector = (a1, a2) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 3t

y = 5 + 2t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

A line through a point (x1,y1) that is parallel to the vector = (a1, a2) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 4 - 7t

y = -3 + 3.5t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = 5t - 3

In the equation y = -3x +1.5, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = -3t +1.5

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 6:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 4:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

When separating a vector into its component form, we are essentially creating a right triangle with the vector being the hypotenuse.

Therefore, we can find each component using the cos (for the x component) and sin (for the y component) functions:

We can now represent these two components together using the denotations i (for the x component) and j (for the y component).

In vector addition, you simply add each component of the vectors to each other.

x component: .

y component: .

z component: .

The new vector is

.

To find the magnitude we use the formula,

Thus its magnitude is 5.

To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

As you set up the equation for y, use the value g = -32.

Finally, find x and y when t = .05:

Use a calculator to solve, making sure you are in degree mode:

This means that after 0.5 seconds, the ball has travelled 17.5 feet horizontally and 5.7 feet vertically.

To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

Write the formula to find both the x and y-components of a vector.

Substitute the value of velocity and theta into the equations.

The vector is:

In order to find the horizontal component, set up an equation involving cosine with 7 as the hypotenuse, since the side in the implied triangle that represents the horizontal component is adjacent to the 22-degree angle:

First, find the cosine of 22, then multiply by 7

To find the vertical component, set up an equation involving sine, since the side in the implied triangle that represents the vertical component is opposite the 22-degree angle:

First, find the sine of 22, then multiply by 7

We are almost done, but we need to make a small adjustment. The picture indicates that the vector points up and to the left, so the horizontal component, 6.49, should be negative: