Find the vector $\mathbf { Y }$ having length $| \mathrm { v } | = 2$
, and direction $\theta = 135 ^ { \circ }$
A) $\mathrm { v } = \langle - 1,1 \rangle$
B) $\boldsymbol { v } = \langle 1 , \sqrt { 2 } \rangle$
C) $\mathrm { v } = \langle - \sqrt { 2 } , \sqrt { 2 } \rangle$
D) $\mathbf { v } = \langle 1,1 \rangle$
E) $\mathrm { v } = \langle - 2,2 \rangle$

Question

Quizplus · Accepted Answer

To find the vector $\mathbf{Y}$ with length $| \mathrm{v} | = 2$ and direction $	heta = 135 ^{\circ}$, we need to first find the unit vector in the direction of $	heta$ and then scale it up by a factor of 2.

The unit vector in the direction of $	heta = 135 ^{\circ}$ can be found as follows:

$\cos 	heta = \cos 135^{\circ} = -\frac{\sqrt{2}}{2}$ and $\sin 	heta = \sin 135^{\circ} = \frac{\sqrt{2}}{2}$

So the unit vector in the direction of $	heta = 135^{\circ}$ is

$\mathbf{u} = \langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} angle$

Scaling this unit vector by a factor of 2 gives us

$\mathbf{Y} = 2 \mathbf{u} = \boxed{	extbf{(C)}\ \langle -\sqrt{2}, \sqrt{2} angle}$

Find the Vector Y\mathbf { Y }Y Having Length ∣v∣=2| \mathrm { v } | = 2∣v∣=2

Find the Vector $\mathbf { Y }$ Having Length $| \mathrm { v } | = 2$