Problem 49 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle- 12,5\rangle $$

Step-by-Step Solution

Answer

Magnitude: 13, Direction: approximately 157.38°

Step 1: Understanding the Vector Components

The vector \( \mathbf{v} \) is given as \( \langle -12, 5 \rangle \). Here, \(-12\) is the \(x\)-component, and \(5\) is the \(y\)-component. We need to determine both the magnitude and direction of this vector.

Step 2: Calculating the Magnitude

The magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) is calculated using the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \). For our vector:\[ \| \mathbf{v} \| = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]Thus, the magnitude of the vector is 13.

Step 3: Calculating the Direction

The direction of a vector is given by the angle \( \theta \) with respect to the positive \(x\)-axis, which can be calculated using the tangent function: \( \tan \theta = \frac{b}{a} \). For our vector \( \mathbf{v} = \langle -12, 5 \rangle \):\[ \tan \theta = \frac{5}{-12} \]To find \( \theta \), take the arctangent:\[ \theta = \arctan \left( \frac{5}{-12} \right) \approx -22.62^{\circ} \]Since the vector is in the second quadrant (negative \(x\)-component and positive \(y\)-component), the angle with respect to the positive \(x\)-axis should be adjusted:\[ \theta = 180^{\circ} + (-22.62^{\circ}) = 157.38^{\circ} \]Thus, the direction of the vector is approximately \(157.38^{\circ}\) from the positive \(x\)-axis.

Step 4: Conclusion

The magnitude of the vector \( \mathbf{v} = \langle -12, 5 \rangle \) is 13, and its direction is approximately \(157.38^{\circ}\) from the positive \(x\)-axis.

Key Concepts

Vector Components

Understanding vector components is crucial when working with vectors. A vector in two dimensions can be expressed in terms of its horizontal and vertical components. For the vector \( \mathbf{v} = \langle -12, 5 \rangle \), the components are as follows:

• \(-12\) is the \(x\)-component. It indicates a movement 12 units in the negative direction along the horizontal axis.
• \(5\) is the \(y\)-component. It represents a movement 5 units in the positive direction along the vertical axis.

These components are crucial, as they help us find further properties of the vector like its magnitude and direction. Basically, they tell you where the vector points and how far it stretches in each dimension. It's like understanding a location on a map by its longitude (\(x\)-value) and latitude (\(y\)-value).

Magnitude of a Vector

The magnitude of a vector gives us the length or size of the vector. It tells us "how much" there is of the vector. Think of it as the distance from the starting point of the vector to its endpoint. To find the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \), you use the formula:\[\| \mathbf{v} \| = \sqrt{a^2 + b^2}\]For the vector \( \mathbf{v} = \langle -12, 5 \rangle \), this becomes:\[\| \mathbf{v} \| = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\]Thus, the magnitude of this vector is 13. This makes it easy to compare it with other vectors or to combine it with others in vector addition. It’s like knowing how far you have to travel on a straight path from start to finish.

Direction of a Vector

The direction of a vector is the angle it forms with the positive \(x\)-axis. It's important because it tells us the orientation of the vector in the plane.We use the tangent function to calculate this angle \( \theta \):\[\tan \theta = \frac{b}{a}\]For \( \mathbf{v} = \langle -12, 5 \rangle \):\[\tan \theta = \frac{5}{-12}\]The direction angle \( \theta \) can then be found using the arctangent function:\[\theta = \arctan\left(\frac{5}{-12}\right)\]However, since \( \theta \) comes out to \(-22.62^{\circ}\), we understand from the vector components that it's in the second quadrant, requiring an adjustment:\[\theta = 180^{\circ} + (-22.62^{\circ}) = 157.38^{\circ}\]Thus, the angle from the positive \(x\)-axis is approximately \(157.38^{\circ}\). This lays out the path the vector takes relative to the \(x\)-axis, kind of like a compass pointing in a specific direction.

Arctangent Function

The arctangent function, abbreviated as \( \tan^{-1} \) or \( \arctan \), is used to find the angle whose tangent value is a given number. It's the inverse function of the tangent, solving the equation \( \tan \theta = x \) for \( \theta \). When applied to our vector \( \mathbf{v} = \langle -12, 5 \rangle \)'s components, we used the arctangent to find:\[\theta = \arctan\left(\frac{5}{-12}\right) \approx -22.62^{\circ}\]The result of an arctangent operation can be tricky because it doesn't directly give us the angle in terms of standard position on a circle. Often, adjustments are necessary to ensure the correct quadrant. In our case, the initial \(-22.62^{\circ}\) was adjusted to \(157.38^{\circ}\) because vectors with a negative \(x\)-component and a positive \(y\)-component lie in the second quadrant. This adjustment guarantees that the direction accurately reflects the real vector orientation in the coordinate plane.