Question

What is the sum V of the twelve vectors that go from the center of a clock to
the hours 1:00,2:00,..., 12:00?
(b) If the 2:00 vector is removed, why do the 11 remaining vectors add to 8:00?
(c) What are the components of that 2:00 vector v = (cos 6, sin 0)?

Answer 1

i can`t understand the concept

Answer 2

For part (a), consider the center of the clock to be the origin. Each vector has an equal magnitude, but opposite in direction vector (12 is opposite 6, 1 is opposite 7, etc.). Adding two such opposing vectors equals zero.\[\left[\begin{matrix}1 \\ 0\end{matrix}\right] + \left[\begin{matrix}-1 \\ 0\end{matrix}\right] = \left[\begin{matrix}0 \\ 0\end{matrix}\right]\] So all 12 vector, 6 pairs, add to zero.

For part (b), if you remove 2 o'clock vector then you are simply left with its opposite vector which no longer is cancelled out. That is the 8 o'clock vector. The other ten vectors still add to zero as they are paired with opposite vectors.

For part (c), it is a question of trigonometry. For a unit circle with a total of 2pi radians, the 3 o'clock vector lies on the 0 radians angle by definition. From 3 to 2 o'clock is 1/12 of the unit circle, or pi/6 radians which gives :\[\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\cos(\pi/6) \\ \sin(\pi/6)\end{matrix}\right] = \left[\begin{matrix}\sqrt{3}/2 \\ 1/2\end{matrix}\right]\] for the components of the 2 o'clock vector.

Answer 3

Thanks man. I thought the same exact concept