Understanding Vector Cross Product and Angle Calculation

In this article, we will delve into the concepts of vector cross product and how to calculate the angle between vectors. We will break down the steps and provide a comprehensive explanation with practical examples.

Introduction to Vector Cross Product

The cross product of two vectors, denoted as mathbf{c} times mathbf{d}, is a vector that is perpendicular to both mathbf{c} and mathbf{d}. This operation is fundamental in vector algebra and has numerous applications in physics, engineering, and computer graphics.

Cross Product Calculation

Given two vectors mathbf{c} 2mathbf{i} - mathbf{j} and mathbf{d} mathbf{i} 2mathbf{j} - 3mathbf{k}, we will calculate their cross product step by step.

Step-by-step Calculation

1. **Set up the determinant**:

[ mathbf{c} times mathbf{d} begin{vmatrix} mathbf{i} mathbf{j} mathbf{k} 2 -1 0 1 2 -3 end{vmatrix} ]

2. **Expand the determinant**:

[ mathbf{c} times mathbf{d} mathbf{i} begin{vmatrix} -1 0 2 -3 end{vmatrix} - mathbf{j} begin{vmatrix} 2 0 1 -3 end{vmatrix} mathbf{k} begin{vmatrix} 2 -1 1 2 end{vmatrix} ]

3. **Calculate the 2x2 determinants**:

begin{vmatrix} -1 0 2 -3 end{vmatrix} (-1)(-3) - (0)(2) 3 begin{vmatrix} 2 0 1 -3 end{vmatrix} (2)(-3) - (0)(1) -6 begin{vmatrix} 2 -1 1 2 end{vmatrix} (2)(2) - (-1)(1) 4 1 5

4. **Combine the results**:

[ mathbf{c} times mathbf{d} 3mathbf{i} - (-6)mathbf{j} 5mathbf{k} 3mathbf{i} 6mathbf{j} 5mathbf{k} ]

Hence, the cross product of mathbf{c} and mathbf{d} is:

[ mathbf{c} times mathbf{d} 3mathbf{i} 6mathbf{j} 5mathbf{k} ]

Finding the Angle Between the Vectors

Using the Dot Product and Magnitudes

The angle between two vectors can be found using the dot product and magnitudes. The formula is:

[ cos theta frac{mathbf{c} cdot mathbf{d}}{|mathbf{c}| |mathbf{d}|} ]

Step-by-step Calculation

1. **Calculate the dot product** mathbf{c} cdot mathbf{d}:

[ mathbf{c} cdot mathbf{d} (2mathbf{i} - mathbf{j}) cdot (mathbf{i} 2mathbf{j} - 3mathbf{k}) 2(1) (-1)(2) 0(-3) 2 - 2 0 ]

2. **Calculate the magnitudes of the vectors**:

|mathbf{c}| sqrt{(2^2 (-1)^2 0^2)} sqrt{4 1} sqrt{5} |mathbf{d}| sqrt{(1^2 2^2 (-3)^2)} sqrt{1 4 9} sqrt{14}

3. **Calculate the angle**:

[ cos theta frac{0}{sqrt{5} cdot sqrt{14}} 0 ]

[ theta cos^{-1}(0) 90^circ ]

Therefore, the angle between vectors mathbf{c} and mathbf{d} is 90 degrees, indicating they are orthogonal.

Summary

The cross product of vectors mathbf{c} and mathbf{d} is 3mathbf{i} 6mathbf{j} 5mathbf{k}, and the angle between them is 90 degrees.