Question

The angle between the vectors a=2i+3j and b=6i-4j is

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given two vectors 'a' and 'b', find the angle between them.

Explanation:

• Let there be two vectors denoted by 'A' and 'B' having an angle $\theta$between them.  
• Then the measure of the angle $\theta$ is given by, $\theta=cos^{-1}(\frac{\vec A.\vec B}{|\vec A||\vec B|})$  
• Now here we have, $\vec a=2\hat i+3\hat j,\ \ \ \vec b=6\hat i-4\hat j$  
• Now the magnitudes of both the vectors are given by,
•                $->|\vec a|=\sqrt{2^2+3^2}=\sqrt{13} \\->|\vec b|=\sqrt{6^2+4^2}=2\sqrt{13}$      
• Now to evaluate the dot product of given two vectors we get,
•               $->\vec a.\vec b=(2\hat i+3\hat j).(6\hat i-4\hat j)\\->\vec a.\vec b=12-12\ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \hat i.\hat i=\hat j.\hat j=1,\ ->\hat i.\hat j=\hat j.\hat i=0) \\->\vec a.\vec b=0$  
• Now putting these values of magnitudes and dot product in the formula for the angle between them we get,
•               $->\theta=cos^{-1}(\frac{0}{2\sqrt{13}(\sqrt{13} ) } )\\->\theta=cos^{-1}(0)\\->\theta=90\ deg.$  
• Hence the angle between the given vectors is $90$°.