Find the angle between the vectors \(\vec{a}\) =  \(\hat{i}\) + 2\(\hat{j}\)+ 3\(\hat{k}\) and \(\vec{b}\) =  2\(\hat{i}\) + \(\hat{j}\)+ 3\(\hat{k}\).

cos^-1(13/14)

We have the vectors \(\vec{a}\) =  \(\hat{i}\) + 2\(\hat{j}\)+ 3\(\hat{k}\) and \(\vec{b}\) =  2\(\hat{i}\) + \(\hat{j}\)+ 3\(\hat{k}\).

Magnitude \(\vec{a}\) = |\(\vec{a}\)| =  √{(1)²+(2)²+(3)²} = √14

Magnitude \(\vec{b}\)  =|\(\vec{b}\)| =  √{(2)²+(1)²+(3)²} = √14

\(\vec{a}\).\(\vec{b}\)= (\(\hat{i}\) + 2\(\hat{j}\)+ 3\(\hat{k}\)).(2\(\hat{i}\) + \(\hat{j}\)+ 3\(\hat{k}\))

\(\vec{a}\).\(\vec{b}\) = ( 2+2+9) =13

we know that \(\vec{a}\).\(\vec{b}\) = |\(\vec{a}\)|.|\(\vec{b}\)| cos(θ)

⇒ cos(θ) = 13/( √14). (√14)

⇒ θ  = cos^-1(13/14)