Question:
Let $\vec{a}=2 \vec{\imath}+3 \vec{\jmath}+4 \vec{k}$ and $\vec{b}=3 \vec{\imath}+4 \vec{\jmath}+5 \vec{k}$.
Find the angle between them.
Solution:
$a=2 i+3 j+4 k$ and $b=3 i+4 j+5 k$
Let angle between them is $\theta$
Then a.b $=2.3+3.4+4.5=38$
$|a|=\sqrt{\left(2^{2}+3^{2}+4^{2}\right)=\sqrt{29}}$
$|a|=\sqrt{\left(2^{2}+3^{2}+4^{2}\right)=\sqrt{29}}$
Now $\cos \theta=a \cdot b /|a||b|=38 / \sqrt{1450}$
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