Finding Angle Between Vectors Using Dot Product

This solution explains how to find the angle between two given vectors using the dot product method. We are given two vectors:

• Vector $\mathbf{a} = 3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}$
• Vector $\mathbf{b} = 4\mathbf{i} - 3\mathbf{j} + \mathbf{k}$

The angle $\theta$ between two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ can be calculated using the dot product formula:

$$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} $$

We need to calculate the dot product $\mathbf{a} \cdot \mathbf{b}$ and the magnitudes $|\mathbf{a}|$ and $|\mathbf{b}|$.

Calculating the Dot Product of Vectors

The dot product (or scalar product) of two vectors $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$ is calculated as:

$$ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 $$

For the given vectors:

• $\mathbf{a} = \langle 3, 2, -6 \rangle$
• $\mathbf{b} = \langle 4, -3, 1 \rangle$

Substituting the components:

$$ \mathbf{a} \cdot \mathbf{b} = (3 \times 4) + (2 \times -3) + (-6 \times 1) $$ $$ \mathbf{a} \cdot \mathbf{b} = 12 + (-6) + (-6) $$ $$ \mathbf{a} \cdot \mathbf{b} = 12 - 6 - 6 $$ $$ \mathbf{a} \cdot \mathbf{b} = 0 $$

Calculating Vector Magnitudes

The magnitude of a vector $\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}$ is given by the formula:

$$ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} $$

Calculate the magnitude of vector $\mathbf{a}$:

$$ |\mathbf{a}| = \sqrt{3^2 + 2^2 + (-6)^2} $$ $$ |\mathbf{a}| = \sqrt{9 + 4 + 36} $$ $$ |\mathbf{a}| = \sqrt{49} $$ $$ |\mathbf{a}| = 7 $$

Calculate the magnitude of vector $\mathbf{b}$:

$$ |\mathbf{b}| = \sqrt{4^2 + (-3)^2 + 1^2} $$ $$ |\mathbf{b}| = \sqrt{16 + 9 + 1} $$ $$ |\mathbf{b}| = \sqrt{26} $$

Determining the Angle Between Vectors

Now, use the dot product formula to find $\cos \theta$:

$$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} $$

Substitute the calculated values:

$$ \cos \theta = \frac{0}{7 \times \sqrt{26}} $$

Since the denominator ($7 \times \sqrt{26}$) is not zero, the result is:

$$ \cos \theta = 0 $$

To find the angle $\theta$, we find the inverse cosine (arccos) of 0:

$$ \theta = \arccos(0) $$ $$ \theta = 90^\circ $$

The angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is $90^\circ$. This means the vectors are perpendicular (orthogonal) to each other.