<p>Hello everyone, can you help me with this question?</p>
<p>Two vectors 2a(vector)+ b(vector) and a(vector)-3b(vector) are perpendicular. find the angle between a(vector) and b(vector),
if a(magnitude only)= 2b(magnitude only)</p>
<p>Thanks for your help,</p>
<p>sry dont know</p>
<p>Let me see...
Since [2a(vector)+b(vector)] is perpendicular to[a(vector)-3b(vector)], we can infer that [2a(vector)+b(vector)]x[a(vector)-3b(vector)] =0
=> 2a^2 - 5a(vector)<em>b(vector) +3b^2 =0.
=> 5a(vector)</em>b(vector) = 2a^2 - 3b^2 = 8b^2-3b^2= 5b^2 (since a=2b)
=> a(vector)*b(vector) = b^2.</p>
<p>Ok, next we have cosine of the angle between a(vector) and b(vector) equals: [a(vector)<em>b(vector)]/[a(magnitude)</em>b(magnitude)]
=b^2/(a<em>b) = b^2/(2b</em>b) =1/2
Therefore, the angle between a(vector) and b(vector) is 60.</p>
<p>p/s: i'm doing this at midnight so my brain may not work well. Correct me if i'm wrong...</p>
<p>I got 60 too. Just like what loser has done.I couldn't figure out another way to do it. is 60 the correct answer btw?</p>