Question

Let \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} .\) Let \(\overrightarrow{\mathbf{c}}=\) \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\). ( \(a\) ) Find \(\mathbf{c}\), expressed in unit vector notation. ( \(b\) ) Find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\).

Short answer

The resultant vector '\(\overrightarrow{\mathbf{c}}\)' after cross product operation is \((-11)\hat{\mathbf{i}} -2\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\). The angle between vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is around 55.41 degrees.

Step-by-step solution

Cross product of vector 'a' and 'b'

The cross product of two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is given by \(c_x = a_y*b_z - a_z*b_y\), \(c_y = a_z*b_x - a_x*b_z\), \(c_z = a_x*b_y - a_y*b_x\). Here, \(c_x\), \(c_y\), \(c_z\) are components of the resulted vector '\(\overrightarrow{\mathbf{c}}\)' in respective directions. By plugging in the given vectors, we get \(\overrightarrow{\mathbf{c}}=(-9 - 2)\hat{\mathbf{i}} + (6-8)\hat{\mathbf{j}} + (4 + 6)\hat{\mathbf{k}}\).

Calculate resultant vector '\(\overrightarrow{\mathbf{c}}\)'

This step involves simple arithmetic calculation for each component of '\(\overrightarrow{\mathbf{c}}\)'. We get \(\overrightarrow{\mathbf{c}}=(-11)\hat{\mathbf{i}} + (-2)\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\).

Calculate dot product and Magnitude of vectors 'a' and 'b'

To find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\), we need to calculate dot product and magnitude. The dot product of two vectors is given by \(a_x*b_x + a_y*b_y + a_z*b_z\), and the magnitude of a vector is given by \(\sqrt{{x^2 + y^2 + z^2}}\). Thus, for our vectors, \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = (2*4)+(-3*-2)+(1*-3) = 8 + 6 -3 = 11\), also \(\|\overrightarrow{\mathbf{a}}\| = \sqrt{{2^2 + (-3)^2 + 1^2}} = \sqrt{4 + 9 +1} = \sqrt{14}\) and \(\|\overrightarrow{\mathbf{b}}\| = \sqrt{{4^2 + (-2)^2 + (-3)^2}} = \sqrt{16+4+9} = \sqrt{29}\).

Calculate the angle between vectors

The angle \(\theta\) between vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) can be calculated as \(\cos(\theta) = \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}} {\|\overrightarrow{\mathbf{a}}\|\*\|\overrightarrow{\mathbf{b}}\|}\). Substituting the calculated values, we get \(\cos(\theta) = \frac{11}{\sqrt{14} * \sqrt{29}} = 0.5785\) and \(\theta = \cos^{-1}(0.5785) = 55.41^\circ\).

Key concepts

Cross Product

The cross product, also known as the vector product, is an operation that takes two vectors and produces another vector. This resulting vector is perpendicular to the plane formed by the original two vectors. The cross product of two vectors \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \) can be calculated using the formula:

• \[ c_x = a_y \cdot b_z - a_z \cdot b_y \]
• \[ c_y = a_z \cdot b_x - a_x \cdot b_z \]
• \[ c_z = a_x \cdot b_y - a_y \cdot b_x \]

These equations allow us to determine each component of the resulting vector \( \overrightarrow{\mathbf{c}} \). The cross product is a fundamental concept in physics and engineering for calculating moments and torques.
For the given vectors \( \overrightarrow{\mathbf{a}} = 2 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + \hat{\mathbf{k}} \) and \( \overrightarrow{\mathbf{b}} = 4 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}} \), the cross product results in:

• \( \overrightarrow{\mathbf{c}} = (-11) \hat{\mathbf{i}} + (-2) \hat{\mathbf{j}} + 10 \hat{\mathbf{k}} \)

Dot Product

The dot product, or scalar product, of two vectors produces a scalar value. Unlike the cross product, this does not result in another vector but a single number. The dot product is calculated using:

• \[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = a_x \cdot b_x + a_y \cdot b_y + a_z \cdot b_z \]

It measures how much two vectors "align" with each other. A dot product of zero indicates perpendicular vectors, whereas a positive or negative value indicates the degree and direction of alignment.
Given vectors in the example, the calculation yields:

• \( \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = 11 \)

The dot product plays a crucial role in determining angles between vectors and in physics to project forces.

Angle Between Vectors

The angle between two vectors can be found using the dot product and magnitudes of the vectors. The relation is given by:

• \[ \cos(\theta) = \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}} {\|\overrightarrow{\mathbf{a}}\|\cdot\|\overrightarrow{\mathbf{b}}\|} \]

\( \theta \) represents the angle between the two vectors. Calculating \( \theta \) involves finding the inverse cosine of the quotient calculated from the dot product and the product of magnitudes.
In our scenario, magnitudes of \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \) are:

• \( \|\overrightarrow{\mathbf{a}}\| = \sqrt{14} \)
• \( \|\overrightarrow{\mathbf{b}}\| = \sqrt{29} \)

Finally, the angle \( \theta \) is approximately \( 55.41^\circ \), showing how much the two vectors are inclined towards each other. This angle is useful in various applications, like navigation or understanding physical phenomena in motion analysis.