# Will a vector at 45° be larger or smaller than its horizontal and vertical components?

Mar 10, 2018

It will be larger

#### Explanation:

A vector at 45 degrees is the same thing as the hypotenuse of an isosceles right triangle.

So, assume you have a vertical component and a horizontal component each of one unit. By the Pythagorean Theorem, the hypotenuse, which is the magnitude of your 45 degree vector will be

$\sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

$\sqrt{2}$ is approximately 1.41, so the magnitude is larger than either the vertical or horizontal component

Mar 14, 2018

Larger

#### Explanation:

Any vector that is not parallel with one of the independent reference (basis) vectors (often, but not always, taken to lie on the x and y axes in the Euclidean plane, particularly when introducing the idea in a mathematics course) will be larger than its component vectors because of the triangle inequality.

There is a proof in the famous book "Euclid's Elements" for the case of vectors in the two dimensional (Euclidean) plane.

So, taking the positive x and y axes as the respective directions of the horizontal and vertical components:

The vector at 45 degrees is not parallel with either the x or the y axis. Therefore, by the triangle inequality, it is larger than either of its components.