Find $| \mathrm { v } |$ and $| \mathbf { u } + \mathbf { y } |$, given that $\mathbf { u } = - \mathbf { j }$ and $\mathbf { v } = \mathbf { i }$
A) $| \mathrm { v } | = 1$
, $| \mathbf { u } + \mathbf { v } | = \sqrt { 2 }$

B) $| v | = - 1$
, $| \mathbf { u } + \mathbf { v } | = - \sqrt { 2 }$

C) $| \mathrm { v } | = \sqrt { 2 }$
, $| \mathbf { u } + \mathbf { y } | = 1$

D) $| \mathrm { v } | = \sqrt { 2 }$
, $| \mathbf { u } + \mathbf { v } | = \sqrt { 2 }$

E)none

Question

Find $| \mathrm { v } |$ and $| \mathbf { u } + \mathbf { y } |$, given that $\mathbf { u } = - \mathbf { j }$ and $\mathbf { v } = \mathbf { i }$
A) $| \mathrm { v } | = 1$
, $| \mathbf { u } + \mathbf { v } | = \sqrt { 2 }$

B) $| v | = - 1$
, $| \mathbf { u } + \mathbf { v } | = - \sqrt { 2 }$

C) $| \mathrm { v } | = \sqrt { 2 }$
, $| \mathbf { u } + \mathbf { y } | = 1$

D) $| \mathrm { v } | = \sqrt { 2 }$
, $| \mathbf { u } + \mathbf { v } | = \sqrt { 2 }$

E)none

Quizplus · Accepted Answer

$|\mathrm{v}|=|\mathbf{i}|=1$, as $\mathbf{i}$ is a unit vector in the x-direction.
$|\mathbf{u}+\mathbf{v}|=|\mathbf{u}+\mathbf{i}|=|-1+\mathrm{i}|=\sqrt{(-1)^2+1^2}=\sqrt{2}$ by Pythagoras' theorem.

Find ∣v∣| \mathrm { v } |∣v∣ And ∣u+y∣| \mathbf { u } + \mathbf { y } |∣u+y∣

Find $| \mathrm { v } |$ And $| \mathbf { u } + \mathbf { y } |$