Angle Between Two Vectors

The angle formed between two vectors is referred to as the angle between the vectors formed at their tails. It can be determined using the dot product (scalar product) or the cross product (vector product). The dot product method involves the cosine of the angle, while the cross-product method uses the sine of the angle and the magnitude of the resultant vector.

1.0What are Vectors?

Vectors are fundamental Mathematical objects used to represent quantities that have both magnitude and direction. Unlike scalars, which only have magnitude, vectors are essential in various fields such as physics, computer science, engineering, and mathematics.

2.0Find Angle Between Two Vectors

To find the angle between two vectors, we rely on the dot product and the magnitudes of the vectors. Let's denote two vectors as $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$.

Dot Product

The dot product (or scalar product) of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is defined as:

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

3.0Angle Between Two Vectors Formula

Using the dot product and magnitudes, the cosine of the angle θ is: $\cos (\theta )=\frac{\overrightarrow{\mathbf{A}}\cdot \overrightarrow{\mathbf{B}}}{|\overrightarrow{\mathbf{A}}||\overrightarrow{\mathbf{B}}|}$

Finally, the angle θ can be found using the inverse cosine function: $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{\mathbf{A}}\cdot \overrightarrow{\mathbf{B}}}{|\vec{A}||\vec{B}|}\right)$

4.0Solved Problems on Angle Between Two Vectors

Example 1: Find the angle between two vectors $3\hat{\mathbf{i}}+5\hat{\mathbf{j}}-2\hat{\mathbf{k}}$ and $2\hat{i}-\hat{j}+2\hat{k}$.

Solution:

$\vec{a}=3\hat{i}+5\hat{j}-2\hat{k}$ and $\vec{b}=2\hat{i}-\hat{j}+2\hat{k}$

The dot product is defined as. $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$

$=(3\hat{i}+5\hat{j}-2\hat{k})\cdot (2\hat{i}-\hat{j}+2\hat{k})$

= 6 – 5 – 4 

= –3

The magnitude of vectors is given by

$|\vec{a}|=\sqrt{3^2+5^2+(-2{)}^2}=\sqrt{a+25+4}=\sqrt{38}$

$|\vec{b}|=\sqrt{2^2+(-1{)}^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3$

So, the angle between the two vector is

$\Rightarrow \theta ={\cos }^{-1}\frac{\vec{a}\cdot \vec{b}}{|\mathrm{a}||\mathrm{b}|}$

$\Rightarrow \theta ={\cos }^{-1}\frac{(-3)}{\sqrt{38}\times 3}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{\sqrt{38}}\right)$

$\Rightarrow \theta =\pi -{\cos }^{-1}\left(\frac{-1}{\sqrt{38}}\right)$

Example 2: Find the angle between two vectors $\vec{a}=\hat{i},\vec{b}=\hat{j}$

Solution:

The dot product is defined as $\vec{a}\cdot \vec{b}$

$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=0$

The magnitude of vectors is given by

$|\vec{a}|=\sqrt{1}=1$

$|\vec{b}|=\sqrt{1}=1$

So, the angle between two vectors is

$\theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{0}{1.1}\right)$

$\Rightarrow \theta ={\cos }^{-1}(0)$

$\Rightarrow \theta =\frac{\pi }{2}$

Example 3: Find the angle between two vectors $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+4\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

Solution:

The dot product is defined as $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$

$\vec{a}\cdot \vec{b}$ = 3(–1) + (–2)·4 + 1·2

= –3 – 8 + 2 

= –9

The magnitude of vectors is given by

$|\vec{a}|=\sqrt{3^2+(-2{)}^2+(1{)}^2}=\sqrt{9+4+1}=\sqrt{14}$

$|\vec{b}|=\sqrt{(-1{)}^2+4^2+2^2}=\sqrt{1+16+4}=\sqrt{21}$

So, the angle between two vectors is

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-9}{\sqrt{14}\sqrt{21}}\right)$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-9}{7\sqrt{6}}\right)$

$\Rightarrow \theta \approx 125.6^{\circ }$

Example 4: Find the angle between two vectors $\vec{a}=2\hat{i}+2\hat{j}+\hat{k}(2)$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$.

Solution:

The dot product is defined as $\vec{a}\cdot \vec{b}$

$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$ = 2·1 + 2·1 + 1·1

= 2 + 2 + 1 = 5

The magnitude of vectors is given by

$|\vec{a}|=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$

$|\vec{b}|=\sqrt{1^2+1^2+1^2}=\sqrt{1+1+2}=\sqrt{3}$

So, the angle between two vectors is

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{3\cdot \sqrt{3}}\right)$

So, the angle between the vectors is approximately ${\cos }^{-1}\left(\frac{5}{3\cdot \sqrt{3}}\right).$

Example 5: Find the angle between the vector $5\hat{\mathbf{i}}-\hat{\mathbf{j}}+4\hat{\mathrm{k}}$ and $3\hat{i}+2\hat{j}+6\hat{k}$

Solution:

The dot product is defined as $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$

$\vec{a}=5\hat{i}-\hat{j}+4\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}+6\hat{k}$

$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$ = 5.3 + (–1)·2 + 4·6

= 15 – 2 + 24

The magnitude of vectors is given by

$|\vec{a}|=\sqrt{5^2+(-1{)}^2+4^2}=\sqrt{25+1+16}=\sqrt{42}$

$|\vec{b}|=\sqrt{3^2+2^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7$

So, the angle between two vectors is

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{37}{7\sqrt{42}}\right)$

5.0Practice Problems on Angle Between Two Vectors

1. Find the angle between vectors $\overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-4\hat{\mathrm{k}}$ and $\vec{B}=-\hat{i}+2\hat{j}+2\hat{k}$ .

2. Determine the angle between the angle $\overrightarrow{\mathrm{A}}=4\hat{\mathrm{i}}+\hat{\mathrm{j}}+7\hat{\mathrm{k}}$ and $\vec{B}=2\hat{i}-3\hat{j}+\hat{k}$ .

3. Compute the angle between two vectors $\vec{A}=6\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{B}=\hat{i}+5\hat{j}-\hat{k}$.

4. Find the angle between vectors $\overrightarrow{\mathrm{A}}=-3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $\vec{B}=\hat{i}-\hat{j}.$

5. Calculate the angle between vectors $\overrightarrow{\mathrm{A}}=7\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{B}}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}$

6.0Sample Questions on Angle Between Two Vectors

1. How would you determine the angle between two vectors?

Solution:

The formula to find the angle θ between two vectors $\overrightarrow{\mathrm{A}}$ and $\vec{B}$ is:

$\cos (\theta )=\frac{\overrightarrow{\mathbf{A}}\cdot \overrightarrow{\mathbf{B}}}{|\overrightarrow{\mathbf{A}}||\overrightarrow{\mathbf{B}}|}$

where $A\cdot B$ is the dot product of the vectors, and $|\overrightarrow{\mathrm{A}}|$ and $|\vec{B}|$ are the magnitudes of $\text{ A }$ and $\text{ B }$, respectively. The angle θ can be found using:

$\theta ={\cos }^{-1}\left(\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\overrightarrow{\mathrm{B}}|}\right)$

2. How do you calculate the dot product of two vectors?

Ans: The dot product of two vectors $A=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\mathrm{B}={\mathrm{b}}_1\hat{\mathrm{i}}+{\mathrm{b}}_2\hat{\mathrm{j}}+{\mathrm{b}}_3\hat{\mathrm{k}}$ is calculated as: $\overrightarrow{\mathbf{A}}\cdot \overrightarrow{\mathbf{B}}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

3. What is the range of possible values for the angle between two vectors?

Ans: The angle θ between two vectors can range from $0^{\circ }$ to $180^{\circ }$ (or 0 to  π radians). An angle of 0° indicates that the vectors are pointing in the same direction, while an angle of 180° indicates that they are pointing in opposite directions.

4. What does it mean if the angle between two vectors is 90°?

Ans: If the angle between two vectors is 90° (or $\frac{\pi }{2}$ radians), it means the vectors are perpendicular to each other. This implies that their dot product is zero.