91Ó°ÊÓ

In the world of vectors, understanding vector components is crucial. A vector in 3D space can be broken down into three parts, each part depicting how much of the vector points in the direction of one of the x, y, or z-axes. This breakdown is helpful because it allows us to work with vectors more intuitively. For example, consider the vector described in the exercise, \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \sqrt{2} \hat{\mathbf{k}} \). This vector has:

• A component of 1 in the direction of \( \hat{\mathbf{i}} \) (the x-axis)
• A component of 1 in the direction of \( \hat{\mathbf{j}} \) (the y-axis)
• A component of \( \sqrt{2} \) in the direction of \( \hat{\mathbf{k}} \) (the z-axis)

By knowing these components, calculating the vector's overall behavior, such as its magnitude or direction, becomes more straightforward. Think of it as knowing how much your three friends are pulling a sled in different directions! This is what makes vector components such a powerful concept in physics and engineering.

The magnitude of a vector tells us how long or strong the vector is, irrespective of its direction. It's like a measure of the vector's size. To find the magnitude, or length, of a vector \( \mathbf{v} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}} \), we use the formula \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \). This formula is derived from the Pythagorean theorem and works for any number of dimensions. In our exercise, the given vector \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \sqrt{2} \hat{\mathbf{k}} \) has components \( a = 1 \), \( b = 1 \), and \( c = \sqrt{2} \). Let's calculate its magnitude step-by-step:

• First, calculate the squares of each component: \( 1^2 = 1 \), \( 1^2 = 1 \), and \((\sqrt{2})^2 = 2 \).
• Next, add these squares together: \( 1 + 1 + 2 = 4 \).
• Finally, find the square root: \( \sqrt{4} = 2 \).

So, the magnitude of this vector is 2. Understanding a vector's magnitude is essential in determining the vector's impact in calculations and real-world applications.

Cosine is one of the fundamental trigonometric functions and is a valuable tool for finding angles between vectors. The cosine of an angle \( \theta \) between a vector and any reference axis is a measure of how much the vector points in the direction of that axis. To find the cosine of the angle between our vector \( \mathbf{v} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \sqrt{2} \hat{\mathbf{k}} \) and the z-axis, we use the formula \( \cos \theta = \frac{\text{z-component}}{|\mathbf{v}|} \). Here's how it works:

• The z-component of our vector is \( \sqrt{2} \).
• The magnitude, as we calculated earlier, is 2.
• Putting it together: \( \cos \theta = \frac{\sqrt{2}}{2} \).

Knowing \( \cos \theta \) allows us to determine the angle \( \theta \). Since \( \cos \theta = \frac{\sqrt{2}}{2} \), the angle between the vector and the z-axis is \( 45^{\circ} \). The cosine function thus provides a direct link between the components of a vector and its orientation relative to a given axis.