# The x and y components of a vector r are r_x = 16 m and r_y = -9.0 m, respectively. (a) Find the...

## Question:

The x and y components of a vector {eq}\vec r {/eq} are {eq}\rm r_x = 16 \ m {/eq} and {eq}\rm r_y = -9.0 \ m {/eq}, respectively.

(a) Find the direction of the vector {eq}\vec r {/eq}?

(b) Find the magnitude of the vector {eq}\vec r {/eq}?

(c) Suppose {eq}\rm r_x {/eq} and {eq}\rm r_y {/eq} are doubled, find the direction and the magnitude of the new vector {eq}\vec r {/eq}.

## Magnitude and Direction of Vectors

Oftentimes, a vector is given in component form. In such a case, we need to know how to find the magnitude and direction of the vector. The magnitude uses the Pythagorean theorem, where the length of the hypotenuse is the square root of the sum of the squares of the components:

$$r = \sqrt{r_x^2 + r_y^2} $$

This can be expanded to more than three dimensions:

$$r = \sqrt{r_x^2 + r_y^2 + r_z^2} $$

The angle of a vector in two dimensions is usually expressed as the angle from the positive *x* axis. Again using what we know of right triangles, the tangent of the angle is the ratio of the *y* component to the *x* component:

$$\tan{(\theta)} = \frac{r_y}{r_x} $$

## Answer and Explanation:

We have a vector with components

$$r_x = 16 \; \mathrm{m}\\ r_y = - 9.0 \; \mathrm{m} $$

We want to find the direction and magnitude of this...

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