Question 967880
Find the angle between the vector a=< sqrt(3),-1> and b=<0,11>. Answer in radians.

a=< sqrt(3),-1> and b=<0,11>
<pre><b><font size = 5>

          a&#8729;b
cos(<font face="symbol">q</font>) = ——————  
         &#8741;a&#8741;&#8741;b&#8741;

{{{cos(theta)}}}{{{""=""}}}{{{((sqrt(3))(0)+(-1)(11))/((sqrt( (sqrt(3))^2+(-1)^2))(sqrt(0^2+11^2)))}}}{{{""=""}}}{{{(0-11)/(sqrt(3+1)sqrt(0+121))}}}{{{""=""}}}{{{(-11)/(sqrt(4)sqrt(121))}}}{{{""=""}}}{{{(-11)/(2*11)}}}{{{""=""}}}{{{-1/2}}} 

{{{theta}}}{{{""=""}}}{{{2pi/3}}}

Edwin</pre></b></font>