# Formula for Angle Between Two Vectors

Here, we will find the formula for angle between two vectors in terms of their direction cosines and also in terms of their direction ratios. The angle between two lines is defined as the angle between two vectors parallel to them. So, the results derived for vectors will also be applicalble to lines,

Let’s begin –

## Formula for Angle Between Two Vectors

#### (a) Angle Between Two Vectors in Terms of Direction Cosines

Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors with direction cosines $$l_1$$, $$m_1$$, $$n_1$$ and $$l_2$$, $$m_2$$, $$n_2$$ respectively. Then, the unit vectors $$\hat{a}$$ and $$\hat{b}$$ in the direction of $$\vec{a}$$ and $$\vec{b}$$ respectively are given by

$$\hat{a}$$ = $$l_1\hat{i} + m_1\hat{j} + n_1\hat{k}$$ and,

$$\hat{b}$$ = $$l_2\hat{i} + m_2\hat{j} + n_2\hat{k}$$

Let $$\theta$$ be the angle between $$\vec{a}$$ and $$\vec{b}$$. Then, $$\theta$$ is also the angle between $$\hat{a}$$ and $$\hat{b}$$

$$\therefore$$ cos$$\theta$$ = $$\hat{a}.\hat{b}\over | \hat{a} | | \hat{b} |$$

$$\implies$$ cos$$\theta$$ = $$(l_1\hat{i} + m_1\hat{j} + n_1\hat{k}).(l_2\hat{i} + m_2\hat{j} + n_2\hat{k})\over (1)(1)$$

$$\because$$ $$\hat{a}$$ and $$\hat{b}$$ are unit vectors, therefore, | $$\hat{a}$$ | and | $$\hat{b}$$ | is 1.

$$\implies$$ cos$$\theta$$ = $$l_1l_2 + m_1m_2 + n_1n_2$$

Condition for Perpendicularity : If $$\vec{a}$$ and $$\vec{b}$$ are perpendicular, then

$$l_1l_2 + m_1m_2 + n_1n_2$$ = 0

Condition for Parallelism : If $$\vec{a}$$ and $$\vec{b}$$ are parallel, then

$$l_1\over l_2$$ = $$m_1\over m_2$$ = $$n_1\over n_2$$

#### (b) Angle Between Two Vectors in Terms of Direction Ratios

Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors with direction ratios $$a_1$$, $$b_1$$, $$c_1$$ and $$a_2$$, $$b_2$$, $$c_2$$ respectively. Then,

$$\vec{A}$$ = A vector along (\vec{a}\) = $$a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$$ and,

$$\vec{B}$$ = A vector along $$\vec{b}$$ = $$a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$$

Let $$\theta$$ be the angle between $$\vec{a}$$ and $$\vec{b}$$. Then, $$\theta$$ is also the angle between $$\vec{A}$$ and $$\vec{B}$$

$$\therefore$$ cos$$\theta$$ = $$\vec{A}.\vec{B}\over | \vec{A} | | \vec{B} |$$

$$\implies$$ cos$$\theta$$ = $$(a_1\hat{i} + b_1\hat{j} + c_1\hat{k}).(a_2\hat{i} + b_2\hat{j} + c_2\hat{k})\over |(a_1\hat{i} + b_1\hat{j} + c_1\hat{k})||(a_2\hat{i} + b_2\hat{j} + c_2\hat{k})|$$

$$\implies$$ cos$$\theta$$ = $$a_1a_2 + b_1b_2 + c_1c_2\over \sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 +{c_2}^2}$$

Condition for Perpendicularity : If $$\vec{a}$$ and $$\vec{b}$$ are perpendicular, then

$$a_1a_2 + b_1b_2 + c_1c_2$$ = 0

Condition for Parallelism : If $$\vec{a}$$ and $$\vec{b}$$ are parallel, then

$$a_1\over a_2$$ = $$b_1\over b_2$$ = $$c_1\over c_2$$

Example : Find the angle between the vectors with the direction ratios proportional to 4, -3, 5 and 3, 4, 5.

Solution : We have,

$$\vec{a}$$ = $$4\hat{i} – 3\hat{j} + 5\hat{k}$$ and $$\vec{b}$$ = $$3\hat{i} + 4\hat{j} + 5\hat{k}$$

Let $$\theta$$ is the angle between the given vectors. Then,

cos$$\theta$$ = $$\vec{a}.\vec{b}\over |\vec{a}||\vec{b}|$$

$$\implies$$ cos$$\theta$$ = $$12 – 12 + 25\over \sqrt{16 + 9 + 25} \sqrt{16 + 9 + 25}$$ = $$1\over 2$$

$$\implies$$ $$\theta$$ = $$\pi\over 3$$