Problem 27 - Precalculus : Building Concepts and Connections

Question

Find the magnitude and direction of each of the given vectors. Express the direction as an angle \(\theta\) in standard position, where \(0^{\circ} \leq \theta<360^{\circ},\) to tuo decimal places. $$\mathbf{v}=\langle 1,1.5\rangle$$

Step-by-Step Solution

Answer

The magnitude of the vector is approximately 1.8 and its direction is approximately 56.31 degrees.

Step 1: Calculate the magnitude

The magnitude \(m\) of vector \(\mathbf{v}=\langle 1,1.5\rangle\) is given by the formula \(m=\sqrt{\mathbf{x}^2+\mathbf{y}^2}\). Using \(\mathbf{v}=\langle 1,1.5 \rangle\), the values are \(\mathbf{x}=1\) and \(\mathbf{y}=1.5\). Substituting these values into the formula gives us \(m=\sqrt{1^2+1.5^2}\) which simplifies to \(m=\sqrt{3.25}\).

Step 2: Compute the Result

The square root of 3.25 equals to 1.8 approximately. Hence, the magnitude ?\(m\) is 1.8.

Step 3: Calculate the Direction

The direction of the vector \(\mathbf{v}\) is given by the formula \(\theta=\arctan\left(\frac{\mathbf{y}}{\mathbf{x}}\right)\). Using \(\mathbf{v}=\langle 1,1.5 \rangle\), the values are \(\mathbf{x}=1\) and \(\mathbf{y}=1.5: \(\theta=\arctan\left(\frac{1.5}{1}\right)\}, which simplifies to \(\theta=\arctan(1.5)\).

Step 4: Compute the Result

The arctan of 1.5 equals to approximately 56.31 degrees. Hence the direction \(\theta\) is 56.31. However, because the vector lies in the first quadrant where \(0^{\circ} \leq \theta < 90^{\circ}\), the direction is already expressed correctly and there is no need for any adjustments.

Key Concepts

Magnitude of a Vector

When dealing with vectors, one of the foundational steps is determining its magnitude. The magnitude of a vector, often seen as the vector's 'length,' gives us an idea of its size or extent.
This is crucial, especially when comparing different vectors or understanding how much force or quantity a vector represents.
To find the magnitude of a vector \( \mathbf{v}=\langle x, y \rangle \), we use the formula:

• \( m = \sqrt{x^2 + y^2} \)

This is derived from the Pythagorean theorem, where the vector represents the hypotenuse of a right triangle formed by its components on the coordinate axes. For example, if we take the vector \( \mathbf{v}=\langle 1,1.5\rangle \), substituting the values into the formula, we get \( m = \sqrt{1^2 + 1.5^2} = \sqrt{3.25} \).
When calculated, this results in a magnitude of approximately 1.8. Understanding this gives clarity on how far the vector stretches from the origin in space.

Direction of a Vector

Apart from the size of the vector, understanding its direction is essential as well. Direction tells us where the vector is pointing in the coordinate plane.
In mathematics and physics, this is usually expressed in terms of an angle with reference to the positive x-axis.
To calculate the direction \( \theta \) of a vector \( \mathbf{v}=\langle x, y \rangle \), we use the arctangent function:

• \( \theta = \arctan\left(\frac{y}{x}\right) \)

For our specific vector \( \mathbf{v}=\langle 1,1.5 \rangle \), we find: \( \theta = \arctan\left(\frac{1.5}{1}\right) = \arctan(1.5) \).
This evaluates to approximately 56.31 degrees. This angle tells us that the vector points diagonally upward and to the right, intersecting the x-axis at roughly 56.31 degrees.
It's essential to ensure the calculated angle falls within the appropriate range, depending on the quadrant the vector lies in.

Angle in Standard Position

Expressing the direction as an angle in standard position involves considering the fundamental measurement convention used in mathematics.
The angle in standard position is measured from the positive x-axis, moving counterclockwise, and can vary from 0 to 360 degrees.
When determining which angle corresponds to a vector, especially critical attention must be given to the component signs:

• If \( x > 0 \) and \( y > 0 \), the angle naturally falls in the first quadrant (0 to 90 degrees).
• If \( x < 0 \) and \( y > 0 \), the angle is in the second quadrant (90 to 180 degrees).
• If \( x < 0 \) and \( y < 0 \), it’s the third quadrant (180 to 270 degrees).
• If \( x > 0 \) and \( y < 0 \), it lands in the fourth quadrant (270 to 360 degrees).

In the case of our vector \( \mathbf{v}=\langle 1,1.5 \rangle \), both components are positive, indicating it lies in the first quadrant.
Thus, the angle \( \theta = 56.31 \) degrees is correct and concise, requiring no further adjustment.
Ensuring that the angle is expressed in these terms is crucial for standardized communication and problem-solving.