Question

The vector u has a magnitude ∥u ∥= 12 ft and direction with generalized angle α = 135°. Find the magnitude of its vertical and horizontal components

Answer 1

Answer:

The magnitude of the vertical component of vector u is  $6\sqrt{2}$

The magnitude of the horizontal component of vector u is  $6\sqrt{2}$

Step-by-step explanation:

A vector quantity has, a magnitude and a direction

The horizontal component of vector $u_{x}$ is ║u║ cos α

The vertical component of vector $u_{y}$ is ║u║ sin α

∵ The magnitude of vector u is 12 ft

∵ The measure of angle α is 135°

- Substitute the magnitude and the angle in the rule of

   each component

∴ $u_{x}$ = 12 cos(135)

∴ $u_{y}$ = 12 sin(135)

∵ cos(135) = $-\frac{\sqrt{2}}{2}$

∴ $u_{x}$ = 12 ( $-\frac{\sqrt{2}}{2}$ )

∴ $u_{x}$ =  $-6\sqrt{2}$

- The magnitude of a number means the number without its sign

∴ The magnitude of the horizontal component of vector u is  $6\sqrt{2}$

∵ sin(135) = $\frac{\sqrt{2}}{2}$

∴ $u_{y}$ = 12 ( $\frac{\sqrt{2}}{2}$ )

∴ $u_{y}$ =  $6\sqrt{2}$

∴ The magnitude of the vertical component of vector u is  $6\sqrt{2}$