Question

TIMES PLEASE HELP Find the magnitude of vector u shown above. Pic of asnwers and triangle!!

Answer 1

Answer:

C. $\vert u \vert = \sqrt{29}$

Step-by-step explanation:

The magnitude of $u$ is given by

$\vert u \vert = \sqrt{(\Delta x )^2+(\Delta y )^2}$,

and to evaluate this we just need to find $\Delta x$ and $\Delta y$.

As we see in the graph

$\Delta x = (-4)-1 = -5$       (the difference in x-coordinates of $u$)

and

$\Delta y = (3)- (1)=2$;         (the difference in y-coordinates of $u$)

Therefore,

$\vert u \vert = \sqrt{(\Delta x )^2+(\Delta y )^2} = \sqrt{(-5)^2+(2)^2} \\\\\boxed{ \vert u \vert = \sqrt{29} }$