Expression for the angle between two vectors \(\vec{A}\) and \(\vec{B}\) :

Expression for the angle between two vectors \(\vec{A}\) and \(\vec{B}\) : 

Let \(\theta\) be the angle between two vectors \(\vec{A}\) and \(\vec{B}\).

\(\therefore \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta\)

or

\(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\)

Let \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\) and \(\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\)

\(\therefore \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z\)

\(|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}\) and \(|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2}\)

Using these values in eqn. (i) we get

$\cos \theta = \frac{A_x B_x + A_y B_y + A_z B_z}{|\vec{A}| |\vec{B}|}$

$= \frac{A_x B_x + A_y B_y + A_z B_z}{\sqrt{A_x^2 + A_y^2 + A_z^2} \sqrt{B_x^2 + B_y^2 + B_z^2}}$

or

\(\theta = \cos^{-1} \left[ \frac{A_x B_x + A_y B_y + A_z B_z}{\sqrt{A_x^2 + A_y^2 + A_z^2} \sqrt{B_x^2 + B_y^2 + B_z^2}} \right]\)

Numerical Example : 

Find the angle between two vectors \(\vec{A} = \hat{i} + 2\hat{j} + \hat{k}\) and \(\vec{B} = 2\hat{i} - \hat{j} + \hat{k}\).

SOLUTION. Here, \(\vec{A} = \hat{i} + 2\hat{j} + \hat{k} \text{ ; } \vec{B} = 2\hat{i} - \hat{j} + \hat{k}\)

$ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$

$= 1 \times 2 + 2 \times (-1) + 1 \times 1 = 2 - 2 + 1 = 1$

$|\vec{A}| = \sqrt{(1)^2 + (2)^2 + (1)^2}= \sqrt{6}$

and $|\vec{B}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{6}$

Let \(\theta\) be the angle between \(\vec{A}\) and \(\vec{B}\), then

\(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{1}{\sqrt{6} \times \sqrt{6}} = \frac{1}{6} = 0.1667\)

or

\(\theta = \cos^{-1} (0.1667) = 80.4^\circ\)