<p>I know this is probably extremely easy, but I never used "arccos"...</p>
<p>If arccos( cos x ) = 0 ; 0 is less than or equal to x is less than or equal to pi/2, then x could = </p>
<p>0
pi/6
pi/4
pi/3
pi/2</p>
<p>Likewise, never solved a problem like the following::
If the magnitudes of vectors a and b are 5 and 12 respectively, then the magnitudes of vector (b - a ) could NOT be
5
7
10
12
17</p>
<p>Thanks!</p>
<p>ArcCos( cos(x)) is just x itself, and you are told this = 0. So x=0.</p>
<p>Think of the vectors a and b as the hour & minute hand of a clock. The greatest magnitude of b-a is 12 + 5, when the hands are at approx 2:42; the smallest magnitude is 12 - 5 = 7, when the hands are at approx. 3:17 . So any value in the range 7 to 17 is possible, but 5 isn't.</p>
<p>i was wondering... how do you find the magnitude of vectors in the first place?</p>
<p>isnt it like v = sqrt(v1+v2) or something?</p>