Problem 61 - 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 \(\mathbf{v}\) $$\mathbf{v}=-10 \mathbf{i}+15 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector is 18.03 and the direction angle is \(123.69^{\circ}\)

Step 1: Compute the magnitude of the vector

The formula for finding the magnitude of a vector v = xi + yj is \( \sqrt{x^2 + y^2} \). Substitute x = -10 and y = 15 into the magnitude formula: \( \|v\| = \sqrt{(-10)^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325}= 18.03 \) to the nearest hundredth.

Step 2: Determine the direction angle

The direction angle \(\theta\) can be obtained using the formula \( \theta = \arctan(\frac{y}{x}) \). Here we need to be careful since the x-component is negative and the y-component is positive, hence the vector lies in the second quadrant. So, to get the actual direction we need to add 180 degrees to the result. Substitute the values of x and y again, which yields \( \theta = \arctan(\frac{15}{-10}) \) and after doing the calculation, the result would be \(-56.31\) degrees. Because this is the second quadrant, we will add 180, \( \theta= 180- 56.31 = 123.69^{\circ} \) to the nearest tenth of a degree.

Key Concepts

Magnitude of a Vector

The magnitude of a vector represents its length or size. Imagine a vector as an arrow; its magnitude is how long that arrow is. To find the magnitude of a vector \(\mathbf{v} = x\mathbf{i} + y\mathbf{j}\), you use the formula \( \sqrt{x^2 + y^2} \). This formula is derived from the Pythagorean theorem, as vectors can be visualized in a 2D coordinate plane.

In our example with vector \(\mathbf{v} = -10\mathbf{i} + 15\mathbf{j}\), we find the magnitude by substituting \(x = -10\) and \(y = 15\) into the formula. Calculating this gives us \( \sqrt{(-10)^2 + 15^2} = \sqrt{325} = 18.03\). We round off to the nearest hundredth, so the magnitude of our vector is \(18.03\).

Understanding magnitude helps in grasping how extensive or intensive a vector's presence is in space.

Direction Angle

The direction angle of a vector indicates the angle it forms with the positive x-axis in a 2D plane. This angle is crucial because it tells us which direction the vector points toward. For vector \(\mathbf{v} = x\mathbf{i} + y\mathbf{j}\), the direction angle \(\theta\) can be found using the tangent function: \( \theta = \arctan(\frac{y}{x}) \).

It's important to recognize which quadrant the vector is in, as it affects the calculation. In our exercise, where \(x = -10\) and \(y = 15\), the vector lies in the second quadrant because the x-component is negative, and the y-component is positive. This tells us that our raw calculation may need adjustment.

Initial calculation gives a negative angle, but adding \(180^\circ\) (because the vector is in the second quadrant) corrects the direction to \(123.69^\circ\), rounded to the nearest tenth of a degree.

Arctan Function

The arctan function, or inverse tangent, is key in finding the angle whose tangent is a given number. When you have \(\frac{y}{x}\), the arctan function helps you find the angle \(\theta\) such that \(\tan(\theta) = \frac{y}{x}\).

In our context, calculating \(\arctan(\frac{15}{-10})\) initially provides an angle corresponding to the division of y by x. However, remember that the arctan value must be adjusted according to the vector's quadrant location.

• Use a calculator with an arctan function to determine this angle accurately.
• The range of arctan typically gives you angles between \(-90^\circ\) and \(90^\circ\).
• Don't forget to adjust your angle based on which quadrant the vector falls into to find the true direction.

Quadrants of Vectors

Vectors are positioned within four quadrants on a coordinate plane, each determined by the signs of their \(x\) and \(y\) components. This placement is critical when calculating direction angles.

Here's a quick guide:

• First Quadrant: \(x > 0\), \(y > 0\) - Both components are positive.
• Second Quadrant: \(x < 0\), \(y > 0\) - The x-component is negative, and the y-component is positive.
• Third Quadrant: \(x < 0\), \(y < 0\) - Both components are negative.
• Fourth Quadrant: \(x > 0\), \(y < 0\) - The x-component is positive, and the y-component is negative.

The quadrant significantly influences whether you add or subtract angles to find true direction. For example, a second quadrant vector requires adding \(180^\circ\) to an angle calculated using arctan. Knowing your vector's quadrant helps ensure calculations remain accurate and reflect the vector's actual path.