# Precalculus : Express a Vector in Component Form

## Example Questions

### Example Question #32 : Algebraic Vectors And Parametric Equations

Express the following vector in component form:

Explanation:

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).

### Example Question #2 : Express A Vector In Component Form

Find , then find its magnitude.  and  are both vectors.

Explanation:

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.

### Example Question #3 : Express A Vector In Component Form

Find the component form of the vector with

initial point

and

terminal point .

Explanation:

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

### Example Question #33 : Algebraic Vectors And Parametric Equations

Find the component form of the vector with

initial point

and

terminal point

Explanation:

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

### Example Question #34 : Algebraic Vectors And Parametric Equations

A bird flies 15 mph up at an angle of 45 degrees to the horizontal.  What is the bird's velocity in component form?

Explanation:

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:

### Example Question #6 : Express A Vector In Component Form

Write this vector in component form:

Explanation:

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: