text: # Find the angle between​​ ​​vector A = i + j - 2k and vector B = i + j - k
  
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  Find the angle between $\vec A=\hat i+\hat j-2\hat k \,and\, \vec B=\hat i+\hat j-\hat k$
  
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  We have
  
  $\vec A.\vec B$= | A | | B | cos θ
  
  Here  | A | = $\sqrt{1^2+1^2+(-2)^2}=\sqrt6$
  
  | B | =$\sqrt{1^2+1^2+(-1)^2}=\sqrt3$
  
  $\vec A.\vec B=(\hat i+\hat j-\hat k).(\hat i+\hat j-\hat k)$
  
  = 1 × 1 + 1 × 1 + (-2)(-1)
  
  = 1 + 1 + 2 = 4
  
  ∴   cos θ = $\frac{\vec A.\vec B}{|A||B|}=\frac{4}{\sqrt6\sqrt3}$
  
  cos θ = $\frac{4}{\sqrt{18}}$
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