The new coordinate system has unit vectors i' and j'
Because of the 20 degree counterclockwise rotation,
i = cos 20 i' -sin 20 j'
j = cos 20 j' +sin 20 i'
For vector A,
A = cos 60 i + sin 60 j
Next, just make the substitution for i and j, and you get the same vector in trannsformed coordinates.
Do the same for Vector B
A =
Vector A has a magnitude 12m and is angled at 60 degrees counterclockwise from the positive direction of the x axis of an xy coord. system. Also. Vector B = (12m)i + (8m)j on that same coord system. Rotate the system counterclockwise about the origin by 20 degrees to form an x'y' system. On this new system, what are (a) Vector A and (b) Vector B, both in unit-vector notation?
2 answers
The rotation matrix, R(θ), for a rotation of θ counter-clockwise (CCW) is:
| cosθ -sinθ |
| sinθ cosθ |
Rotation of the basis by 20° CCW is the same as rotating the vectors 20° CW, or θ=-20°.
A = (12cos(60°),12sin(60°))
= (6 m, 6√3 m)
B = (12 m, 8 m)
Vector A in the new reference is therefore
A'
= R(-20°)A
=
| cos(-20°) -sin(-20°) | |6 |
| sin(-20°) cos(-20°) | |6√3|
=
(6cos(-20°)+6√3(-sin(-20°)), 6sin(-20°)+6√3 cos(-20°) )
= (5.64 + 3.55, -2.05+9.77)
= (9.19, 7.71)
(check: √(9.19²+7.71²)=12, OK)
The rotation of vector B' can be worked out similarly.
| cosθ -sinθ |
| sinθ cosθ |
Rotation of the basis by 20° CCW is the same as rotating the vectors 20° CW, or θ=-20°.
A = (12cos(60°),12sin(60°))
= (6 m, 6√3 m)
B = (12 m, 8 m)
Vector A in the new reference is therefore
A'
= R(-20°)A
=
| cos(-20°) -sin(-20°) | |6 |
| sin(-20°) cos(-20°) | |6√3|
=
(6cos(-20°)+6√3(-sin(-20°)), 6sin(-20°)+6√3 cos(-20°) )
= (5.64 + 3.55, -2.05+9.77)
= (9.19, 7.71)
(check: √(9.19²+7.71²)=12, OK)
The rotation of vector B' can be worked out similarly.
0 answers