Problem 64 - Precalculus with Limits

Question

In Exercises 63-66, find the magnitude and direction angle of the vector \(\mathbf{v}\). \(\small{\mathbf{v} = -5\mathbf{i} + 4\mathbf{j}}\)

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{v} = -5\mathbf{i} + 4\mathbf{j}\) is \(\sqrt{41}\), and the direction angle is \(\pi - \arctan(0.8)\).

Step 1: Find the Magnitude of the Vector

The magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) is given by the formula \(||\mathbf{v}|| = \sqrt{a^2 + b^2}\). Using this formula, the magnitude of the vector \(\mathbf{v} = -5\mathbf{i} + 4\mathbf{j}\) is \(\sqrt{(-5)^2 + 4^2} = \sqrt{41}\).

Step 2: Find the Direction Angle of the Vector

The direction angle \(\theta\) of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) is given by the formula \(\theta = \arctan(b/a)\). However, the vector \(\mathbf{v} = -5\mathbf{i} + 4\mathbf{j}\) is in the second quadrant, where the arctan function is not applicable directly. So we use the identity \(\theta = \pi - \arctan(|b/a|)\) for vectors in the second quadrant. This gives us \(\theta = \pi - \arctan(|4/(-5)|) = \pi - \arctan(0.8)\).

Step 3: Calculate the Value of Direction Angle

Now we need to calculate or estimate the value of \(\pi - \arctan(0.8)\). It's usually okay at this point to use a calculator to get the approximate numeric value, if required.

Key Concepts

Magnitude of a Vector

To find the magnitude of a vector is often the first step in understanding its scale and the distance it represents in geometric space.The magnitude tells us how long a vector is, and it's a crucial component in many calculations.For a vector given by \[\mathbf{v} = a\mathbf{i} + b\mathbf{j} \]the magnitude is computed using the Pythagorean theorem as:\[||\mathbf{v}|| = \sqrt{a^2 + b^2}. \]This formula resembles finding the hypotenuse of a right triangle, which gives you a sense of how these components interact geometrically.
For example, if we look at the vector \( \mathbf{v} = -5\mathbf{i} + 4\mathbf{j} \), substitute \( a = -5 \) and \( b = 4 \) into the formula:- Calculate \(-5^2 = 25\)- Calculate \(4^2 = 16\)- Add the squares to get \(25 + 16 = 41\)- The magnitude thus is \( \sqrt{41} \), an approximate value of 6.4.
This value is always positive and specifies how much space the vector covers without considering its direction.

Direction Angle

The direction angle provides orientation of a vector in the coordinate plane and is particularly useful in navigation and physics. To find this angle, which is usually labeled as \( \theta \), we use the tangent function. For the vector \[\mathbf{v} = a\mathbf{i} + b\mathbf{j} \]the direction angle is given by:\[\theta = \arctan\left(\frac{b}{a}\right). \]
Yet, it's crucial to consider the quadrant in which the vector lies, because the tangent function's output usually refers to angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).If a vector is in the second quadrant, like \( \mathbf{v} = -5\mathbf{i} + 4\mathbf{j} \), adjustments are necessary.This particular vector requires the identity adjustment:- Use \( \theta = \pi - \arctan(|\frac{4}{-5}|) = \pi - \arctan(0.8) \).
Utilizing a calculator can help to find that \( \theta \approx 2.5 \) radians.Understanding the angle ensures you know the exact direction in which the vector points.

Vector Components

Every vector in the plane can be broken down into two primary components: the x-component and the y-component.These components reflect how much the vector stretches in the horizontal and vertical directions. Using the standard form of a vector:\[\mathbf{v} = a\mathbf{i} + b\mathbf{j} \]the components are clearly seen as the coefficients of \(\mathbf{i}\) and \(\mathbf{j}\).For the provided vector \( \mathbf{v} = -5\mathbf{i} + 4\mathbf{j} \):- The x-component is -5 (since it pairs with \(\mathbf{i}\))- The y-component is 4 (since it pairs with \(\mathbf{j}\))
These components are essential because they allow you to reconstruct the vector graphically by showing steps which are meant to be taken horizontally, then vertically, starting from an origin point. Furthermore, vector components can help determine the direction and magnitude, acting like building blocks to form the entire vector.