Question

Consider a= ⟨−2,5,1⟩, b= ⟨−5,2,−5⟩ and c= ⟨−1,−5,4⟩. Find the angles between the following vectors: The angle between a and b is:

Answer 1

Answer:

$cos^{-1}(\frac{\sqrt 5}{6})$

Step-by-step explanation:

We are given that

$a=,b=$ and c=<-1,-5,4>

We have to find the angles between  a and b .

$\mid a\mid=\sqrt{(-2)^2+(5)^2+(1)^2}=\sqrt{30}$

$\mid b\mid=\sqrt{(-5)^2+(2)^2+(5)^2}=3\sqrt{6}$

$a\cdot b=\cdot =10+10-5=15$

Angle between two vectors a and b is given  by

$cos\theta=\frac{a\cdot b}{\mid a\mid \mid b\mid }$

Using the formula

$cos\theta=\frac{15}{\sqrt{30}\times 3\sqrt{6}}$

$cos\theta=\frac{5}{6\sqrt{5}}$

$cos\theta=\frac{5\times \sqrt 5}{6\times (\sqrt 5)^2}=\frac{\sqrt 5}{6}$

$\theta=cos^{-1}(\frac{\sqrt 5}{6})$

Hence, the angle between a and b=$cos^{-1}(\frac{\sqrt 5}{6})$