Answer:
The magnitude of the vertical component of vector u is [tex]6\sqrt{2}[/tex]
The magnitude of the horizontal component of vector u is [tex]6\sqrt{2}[/tex]
Step-by-step explanation:
A vector quantity has, a magnitude and a direction
The horizontal component of vector [tex]u_{x}[/tex] is ║u║ cos α
The vertical component of vector [tex]u_{y}[/tex] 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
∴ [tex]u_{x}[/tex] = 12 cos(135)
∴ [tex]u_{y}[/tex] = 12 sin(135)
∵ cos(135) = [tex]-\frac{\sqrt{2}}{2}[/tex]
∴ [tex]u_{x}[/tex] = 12 ( [tex]-\frac{\sqrt{2}}{2}[/tex] )
∴ [tex]u_{x}[/tex] = [tex]-6\sqrt{2}[/tex]
- The magnitude of a number means the number without its sign
∴ The magnitude of the horizontal component of vector u is [tex]6\sqrt{2}[/tex]
∵ sin(135) = [tex]\frac{\sqrt{2}}{2}[/tex]
∴ [tex]u_{y}[/tex] = 12 ( [tex]\frac{\sqrt{2}}{2}[/tex] )
∴ [tex]u_{y}[/tex] = [tex]6\sqrt{2}[/tex]
∴ The magnitude of the vertical component of vector u is [tex]6\sqrt{2}[/tex]
