Find the angle between the vectors \((3\vec{i}+4\vec{j}), (\vec{i}-\vec{j}+\vec{k})\) using their vector Product:

Concept:

Cross/Vector product of two vectors is defined as:

\({\rm{ \vec{A} \times \vec{B} = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times sin}}\;{\rm{\theta }} \times \rm ̂{n}\)

where θ is the angle between \({\rm{⃗ A}}\;{\rm{and}}\;{\rm{⃗ B}}\)

where \(\rm ̂ n\) is a unit vector

If \(\rm \vec A = a_1̂ i +a_2̂ j+ a_3̂ k\) and \(\rm \vec B = b_1̂ i +b_2̂ j+b_3 ̂ k\), then their cross product is:

\(\rm \vec A\times\vec B=\begin{vmatrix} \rm ̂ i & \rm ̂ j & \rm ̂ k \\ \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}\).

Calculation:

Let,

\(\vec{a}\ =\ (3\vec{i}+4\vec{j})\)

\(\vec{b}\ =\ (\vec{i}-\vec{j}+\vec{k})\)

\(\rm \vec a\times\vec b=\begin{vmatrix} \rm ̂ i & \rm ̂ j & \rm ̂ k \\ \rm 3 & \rm 4 & \rm 0 \\ \rm 1 & \rm -1 & \rm 1\end{vmatrix}\)

= î(4 + 0) - ĵ (3 - 0) + k̂(- 3 - 4)

⇒ \((3\vec{i}+4\vec{j})\times (\vec{i}-\vec{j}+\vec{k}) = 4\vec{i}-3\vec{j}-7\vec{k}\)

Now,

 \(|4\vec{i}-3\vec{j}-7\vec{k}|=\ \sqrt{4^2\ +\ 3^2\ +\ 7^2}\)

⇒ \(|4\vec{i}-3\vec{j}-7\vec{k}|=\ \sqrt{74}\)

\(\Rightarrow \ \sqrt{74}\ =|\vec{a}||\vec{b}| \sin \theta\)

\(\Rightarrow \ \sqrt{74}\ = 5\sqrt{3}\sin \theta\)

Therefore, 

 \(\sin \theta = \frac{\sqrt{74}}{5\sqrt{3}}\)