It will be helpful to first draw the vector so we can see what quadrant the angle is in:

Since the vector is pointing up and to the right, it is in the first quadrant. To determine the angle, set up a trig equation with tangent, since the component 5 is opposite and the component 4 is adjacent to the angle we are looking for:

to solve for theta, take the inverse tangent of both sides:

Now we have the direction, and we can solve for the magnitude using Pythagorean Theorem:

take the square root of both sides

The vector in polar form is

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 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:

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.