Question

Vector A (with arrow above) has a negative x component 3.26 units in length and a positive y component 1.94 units in length.

(a) Determine an expression for A (with arrow above) in unit-vector notation.
(__i+__j) units

(b) Determine the magnitude and direction of A(with arrow above).
magnitude_____unit(s)
direction______° counterclockwise from the positive x axis

(c) What vector B(with arrow above) when added to vector A (with arrow above), gives a resultant vector with no x component and a negative y component 4.8 units in length?
(__i+__j) units

Answer 1

a)

x component goes before 'i' while y component goes before 'j'

Answer 2

so it would be xi+yj?

Answer 3

yeah, but you have to put actual numbers

Answer 4

or 3.26i+1.94j

Answer 5

almost
it says
" negative x component"

so it should be

-3.26i+1.94j

Answer 6

ok

Answer 7

to find magnitude , take the square root of sum of square (of x and y)

\[\sqrt{(-3.26)^2+\left(1.94\right)^2}\]

Answer 8

b)magnitude of a=(3sqrt58)/10

Answer 9

i got 3.79

Answer 10

direction would be 1.95/3.26?

Answer 11

yes, magnitude is correct

Answer 12

arc tan?

Answer 13

for direction , you take arctan

Answer 14

\[\text{Arctan}\left(\frac{1.94}{3.26}\right)\]

this will get you a terminal angle

Answer 15

i got 329.24...bc i got a negative and had to add 360

Answer 16

this is the angle we want

Answer 17

ok

Answer 18

we got that angle to be 30.7565