Problem 54 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1. $$\mathbf{w}=\mathbf{i}-2 \mathbf{j}$$.

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\) is \(\mathbf{u} = \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{j}\), and its magnitude has been verified to be 1.

Step 1: Find the magnitude of the given vector

The magnitude (or length) of a vector \(\mathbf{w}\) with components \(w_x\) and \(w_y\) in 2D space is given by the formula \(\|\mathbf{w}\| = \sqrt{{w_x}^2 + {w_y}^2}\). In this case, \(w_x\) = 1 (the coefficient of \(\mathbf{i}\)) and \(w_y\) = -2 (the coefficient of \(\mathbf{j}\)), so the magnitude of the vector \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\) is \(\|\mathbf{w}\| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\).

Step 2: Find the unit vector in the direction of the given vector

The unit vector in the direction of a given vector \(\mathbf{w}\) is obtained by dividing \(\mathbf{w}\) by its magnitude. That is, \(\mathbf{u} = \frac{\mathbf{w}}{\|\mathbf{w}\|}\). So, the unit vector in the direction of \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\) is \(\mathbf{u} = \frac{\mathbf{i} - 2\mathbf{j}}{\sqrt{5}} = \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{j}\)

Step 3: Verify the magnitude of the unit vector

To confirm that the resulting vector is indeed a unit vector, its magnitude must be calculated and shown to be 1. By using the formula \(\|\mathbf{u}\| = \sqrt{{u_x}^2 + {u_y}^2}\), we find that \(\|\mathbf{u}\| = \sqrt{(\frac{1}{\sqrt{5}})^2 + (-\frac{2}{\sqrt{5}})^2} = \sqrt{\frac{1}{5} + \frac{4}{5}} = \sqrt{1} = 1\) Therefore, the vector \(\mathbf{u} = \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{j}\) is indeed a unit vector.

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is similar to finding the length of a line segment in geometry. It's a measure of how long the vector is, regardless of its direction. For a two-dimensional (2D) vector like \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\), you calculate its magnitude using the formula:
\[ \|\mathbf{w}\| = \sqrt{{w_x}^2 + {w_y}^2} \]
where \(w_x\) and \(w_y\) are the components of the vector, in this case, 1 and -2 respectively. Plugging these values in, we find:

• \(w_x = 1\)
• \(w_y = -2\)

\[\|\mathbf{w}\| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\]
This magnitude tells us how far from the origin the vector \(\mathbf{w}\) extends in the coordinate system.

2D Vectors

Two-dimensional vectors are essential in various fields such as physics and engineering because they are the simplest form of vectors that still show direction and magnitude.
In a 2D plane, vectors are expressed using horizontal and vertical components, often represented as \(\mathbf{i}\) (horizontal) and \(\mathbf{j}\) (vertical).
For example, the vector \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\) comprises:

• \(\mathbf{i}\) which moves 1 unit along the horizontal axis.
• \(-2\mathbf{j}\) which moves 2 units in the opposite direction along the vertical axis.

These components help represent any movement in the plane, making calculations like addition, subtraction, and finding resultant vectors straightforward.

Vector Normalization

Vector normalization is a process of converting a non-unit vector to a unit vector, maintaining its direction but adjusting its magnitude to 1.
This is particularly useful when you want a direction without scaling involved.
For a given vector \(\mathbf{w}\), the unit vector \(\mathbf{u}\) is found by dividing each component by the vector's magnitude:
\[ \mathbf{u} = \frac{\mathbf{w}}{\|\mathbf{w}\|} \]
In our example, the vector \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\) has a magnitude of \(\sqrt{5}\). The unit vector it forms is:

• \(\mathbf{u} = \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{j}\)

To verify \(\mathbf{u}\) is a unit vector, we find its magnitude should be precisely 1:
\[\|\mathbf{u}\| = \sqrt{\left(\frac{1}{\sqrt{5}}\right)^2 + \left(-\frac{2}{\sqrt{5}}\right)^2} = \sqrt{\frac{1}{5} + \frac{4}{5}} = \sqrt{1} = 1\]
This ensures \(\mathbf{u}\) is indeed normalized, demonstrating its scaled-down length with preserved direction.