Introduction

In this article, we will explore the concept of vectors and how to find a vector that is parallel to a given vector. We will also delve into how to construct a vector that has a specific magnitude and is parallel to a given vector. This is a common task in vector arithmetic, particularly in fields such as physics and engineering.

Understanding Vector Parallelism

A vector (mathbf{x}) is said to be parallel to a vector (mathbf{A}) if (mathbf{x}) can be expressed as a scalar multiple of (mathbf{A}). This means that there exists a scalar (k) such that (mathbf{x} k mathbf{A}). The scalar (k) can be any real number, which allows for flexibility in determining the vector's magnitude and direction.

Example: Finding Scalar Multiple Vectors

Given:

[mathbf{A} 2mathbf{i} 3mathbf{j} - 1mathbf{k}]

Let's express a vector (mathbf{x}) that is parallel to (mathbf{A}) using a scalar (k). Thus, we have:

[mathbf{x} k mathbf{A} k(2mathbf{i} 3mathbf{j} - 1mathbf{k}) 2kmathbf{i} 3kmathbf{j} - kmathbf{k}]

For example, if (k 1), then:

[mathbf{x} 2mathbf{i} 3mathbf{j} - 1mathbf{k}]

And if (k 2), then:

[mathbf{x} 4mathbf{i} 6mathbf{j} - 2mathbf{k}]

Therefore, any scalar multiple of (mathbf{A}) will yield a vector that is parallel to (mathbf{A}).

Converting a Vector to an Unit Vector

An important concept in vector arithmetic is the unit vector, which is a vector of magnitude 1 that points in the same direction as the original vector. We can convert (mathbf{A}) to a unit vector by dividing each component of (mathbf{A}) by its magnitude. The magnitude of (mathbf{A}) is given by:

[|mathbf{A}| sqrt{a^2 b^2 c^2} sqrt{2^2 3^2 (-1)^2} sqrt{4 9 1} sqrt{14}]

The unit vector (mathbf{A}^) is then:

[mathbf{A}^mathbf{} frac{2mathbf{i} 3mathbf{j} - mathbf{k}}{sqrt{14}} frac{2}{sqrt{14}}mathbf{i} frac{3}{sqrt{14}}mathbf{j} - frac{1}{sqrt{14}}mathbf{k}]

Now, if we want to find a vector (mathbf{X}) that is parallel to (mathbf{A}) and has a specific magnitude, say (2sqrt{6}), we use the unit vector (mathbf{A}^) and scale it accordingly:

[mathbf{X} 2sqrt{6} cdot mathbf{A}^mathbf{} 2sqrt{6} left(frac{2}{sqrt{14}}mathbf{i} frac{3}{sqrt{14}}mathbf{j} - frac{1}{sqrt{14}}mathbf{k}right)]

This can be simplified to:

[mathbf{X} frac{2 sqrt{6} cdot 2}{sqrt{14}}mathbf{i} frac{2 sqrt{6} cdot 3}{sqrt{14}}mathbf{j} - frac{2 sqrt{6} cdot 1}{sqrt{14}}mathbf{k}]

Which further simplifies to:

[mathbf{X} frac{4 sqrt{6}}{sqrt{14}}mathbf{i} frac{6 sqrt{6}}{sqrt{14}}mathbf{j} - frac{2 sqrt{6}}{sqrt{14}}mathbf{k} frac{2 sqrt{6}}{sqrt{21}}mathbf{i} frac{3 sqrt{6}}{sqrt{21}}mathbf{j} - frac{sqrt{6}}{sqrt{21}}mathbf{k}]

Therefore, the vector (mathbf{X}) is:

[mathbf{X} frac{2}{sqrt{21}}mathbf{i} frac{3}{sqrt{21}}mathbf{j} - frac{1}{sqrt{21}}mathbf{k}]

This vector (mathbf{X}) is parallel to (mathbf{A}) and has the desired magnitude.

Conclusion

In this article, we have discussed the concept of vectors and how to find vectors that are parallel to a given vector by using scalar multiples. We have also explored the process of converting a vector to a unit vector and scaling it to achieve a desired magnitude. These techniques are foundational in vector arithmetic and are widely applicable in various scientific and engineering fields.