Problem 62 - Algebra and Trigonometry

Question

In Exercises \(61-64\), find the magnitude livi, to the nearest hundredth, and the direction angle \(\theta,\) to the nearest tenth of a degree, for each given vector \(\mathbf{v}\) $$\mathbf{v}=2 \mathbf{i}-8 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{v} = 2\mathbf{i} - 8 \mathbf{j}\) is approximately 8.25, and the direction angle \(\theta\) is approximately 103.6° above the negative x-axis.

Step 1: Calculate the magnitude of the vector

The magnitude of the vector \(\mathbf{v}\) can be calculated using the formula \(\lvert \mathbf{v} \rvert = \sqrt{(2^2) + (-8^2)}\). Substitute the components of the given vector into the formula and solve.

Step 2: Calculate the direction angle

The direction angle of \(\theta\) can be found using the formula \(\theta = \tan^{-1}\left(\frac{-8}{2}\right)\). This is because the \(\mathbf{i}\) component corresponds to the x-axis, and the \(\mathbf{j}\) component corresponds to the y-axis. Remember that because the y-component is negative, the angle will be in the fourth quadrant, so we need to subtract it from 360°.

Step 3: Use a calculator for the computations

Finally, use a calculator to compute the square root and the inverse tangent to find the magnitude and direction angle, respectively. Note that when calculating the direction angle, the value must be converted to degrees by multiplying by \(\frac{180}{\pi}\).

Key Concepts

Understanding the Magnitude of a Vector

When you think of a vector, imagine an arrow pointing a certain direction. The magnitude of a vector is similar to the length of this arrow. To find the magnitude, we use the formula:

• For vector \( \mathbf{v} = 2 \mathbf{i} - 8 \mathbf{j} \), the magnitude \( \lvert \mathbf{v} \rvert \) is given by \( \sqrt{(2^2) + (-8^2)} \).

Simply put:

• First, square each of the components -- 2 and -8 in this case. So we have \( 2^2 = 4 \) and \( (-8)^2 = 64 \).
• Next, add these squares: \( 4 + 64 = 68 \).
• Finally, take the square root of the sum: \( \sqrt{68} \approx 8.25 \).

This value, 8.25, gives us the length of the vector. It's like measuring the distance from the origin to the point \( (2,-8) \) in a grid.
Understanding this step is crucial, as it provides insight into how far a vector “stretches” in space, which is often used in physics and engineering applications.

Finding the Direction Angle of a Vector

The direction angle tells us which way the vector is pointing. Imagine it as finding which direction an arrow is facing. For our vector \( \mathbf{v} = 2 \mathbf{i} - 8 \mathbf{j} \), its direction is determined by where it stands with respect to the positive x-axis.

• To find the direction angle \( \theta \), we use the formula: \( \theta = \tan^{-1}\left(\frac{-8}{2}\right) \).

This formula comes from the trigonometric tangent function. The \( -8 \) and \( 2 \) represent directions in the vertical and horizontal axes (y and x, respectively).
Now, since the y-component is negative, it indicates the vector is pointing downward. The calculated angle is thus in the fourth quadrant (below the x-axis for a standard coordinate plane). To adjust the angle to make sense with the direction, subtract it from 360°. So:

• Calculate \( \tan^{-1}\left(-4\right) \) to get an initial angle in radians, then convert it to degrees.
• Adjust by computing: \( 360° - \text{calculated angle} \).

That gives us the final direction angle for the vector, indicating the precise direction in which it "points."

Learning About Inverse Trigonometric Functions

Inverse trigonometric functions help us work out angles from known ratio values. They are the reverse of regular trigonometric operations.
For example, when we use \( \tan^{-1} \), what we are doing is finding the angle for a given tangent value. Like this:

• Here, \( \theta = \tan^{-1}\left(\frac{-8}{2}\right) \). So, we want to know "what angle has a tangent of \(-4\)?"

Using a calculator, you can find an angle corresponding to this ratio. Typically, the result is in radians, so conversion might be needed:

• Multiply radians by \( \frac{180}{\pi} \) to convert to degrees.

This feature is especially valuable in a wide range of mathematics applications where angles in right triangles are required. Once you grasp these functions, finding angles for any known side ratios becomes a straightforward calculation.
Inverse trigonometric functions bridge the gap between trigonometric ratios and angles, making them indispensable for angle calculations directly from coordinate data.