Vector A has magnitude 100 N, lies in the xy-plane and has an angle 30 degrees measured from the negative x-axis clockwise. The unit vector expression of this vector is
Answers
Answer:
Answer
Noting that the give 13060 is measured counter clockwise from the +x axis the two vectors can be written as
A
=8.00(cos(130
0
)
i
^
+sin(130
0
)
j
^
)=−5.14
i
^
+6.13
j
^
B
=B
x
i
^
+B
y
j
^
=−7.72
i
^
−9.20
j
^
(a) The angle between the negative direction of the y axis(
−j
^
) and the direction of
A
is
θ=cos
−1
(
A
A
.(
−j
^
)
)=cos
−1
(
(−5.14)
2
+(6.13)
2
−6.13
)=cos
−1
(
8.00
−6.13
)=140
0
Alternatively, one may say that the −y direction corresponds to an angle of 270
0
, and the answer is simply given by 270
0
−130
0
=140
0
.
(b) Since the y axis is inthe xy plane , an
A
×
B
is perpendicular to that plane, then the answer is 90.0
0
(c) The vector can be simplified as
A
×(
B
+3.00
k
^
)=(−5.14
i
^
+6.13
j
^
)×(−7.72
i
^
−9.20
j
^
+3.00
k
^
)
=18.39
i
^
+15.42
j
^
+94.61
k
^
In magnitude is ∣
A
×(
B
+3.00
k
^
)∣=97.6 the angle between the negative direction of the y axis (−
j
^
) and the direction of the above vector is
θ=cos
−1
(
97.6
−15.42
)=99.1
0