Answer:
[tex]cos^{-1}(\frac{\sqrt 5}{6})[/tex]
Step-by-step explanation:
We are given that
[tex]a=<-2,5,1>,b=<-5,2,-5>[/tex] and c=<-1,-5,4>
We have to find the angles between a and b .
[tex]\mid a\mid=\sqrt{(-2)^2+(5)^2+(1)^2}=\sqrt{30}[/tex]
[tex]\mid b\mid=\sqrt{(-5)^2+(2)^2+(5)^2}=3\sqrt{6}[/tex]
[tex]a\cdot b=<-2,5,1>\cdot <-5,2,-5>=10+10-5=15[/tex]
Angle between two vectors a and b is given by
[tex]cos\theta=\frac{a\cdot b}{\mid a\mid \mid b\mid }[/tex]
Using the formula
[tex]cos\theta=\frac{15}{\sqrt{30}\times 3\sqrt{6}}[/tex]
[tex]cos\theta=\frac{5}{6\sqrt{5}}[/tex]
[tex]cos\theta=\frac{5\times \sqrt 5}{6\times (\sqrt 5)^2}=\frac{\sqrt 5}{6}[/tex]
[tex]\theta=cos^{-1}(\frac{\sqrt 5}{6})[/tex]
Hence, the angle between a and b=[tex]cos^{-1}(\frac{\sqrt 5}{6})[/tex]
