Posted by **Rick** on Sunday, June 27, 2010 at 4:27pm.

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?

- Physics -
**drwls**, Sunday, June 27, 2010 at 4:45pm
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 =

- Physics -
**MathMate**, Sunday, June 27, 2010 at 4:56pm
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.

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