Problem 65 - Precalculus with Limits

Question

Finding the Direction Angle of a Vector In Exercises \(63-66\) , find the magnitude and direction angle of the vector v. $$\mathbf{v}=3\left(\cos 60^{\circ} \mathbf{i}+\sin 60^{\circ} \mathbf{j}\right)$$

Step-by-Step Solution

Answer

The magnitude of the vector is approximately 3 and the direction angle is approximately 60 degrees.

Step 1: Analyse the vector

The given vector can be written as \(\mathbf{v}=3\left(\cos 60^{\circ} \mathbf{i}+\sin 60^{\circ} \mathbf{j}\right)\). It appears that the vector is in the polar form, with magnitude 3 and direction angle 60 degrees. Its Cartesian form would derive directly from the given form.

Step 2: Calculate the x and y components of the vector

Given that the direction angle is 60 degrees, we can calculate the x and y components. The x component (i) can be calculated as: \(v_x=3\cos(60^{\circ})=3\cdot \frac{1}{2}=1.5\). The y component (j) is calculated as: \(v_y=3\sin(60^{\circ})=3\cdot \frac{\sqrt{3}}{2}\approx2.6\).

Step 3: Calculate the magnitude

The magnitude of the vector is given by the formula \(\|v\|= \sqrt{{v_x}^2+{v_y}^2}=\sqrt{(1.5)^2+(2.6)^2}\approx 3\), which was the original magnitude.

Step 4: Find the Direction Angle

The direction angle can be found by applying the arctan formula: \(\theta = \arctan{(\frac{v_y}{v_x})} = \arctan{(\frac{2.6}{1.5})}\approx 60^{\circ}\), which was the original direction angle.

Key Concepts

Vector Magnitude

The magnitude of a vector is a measure of its 'length' or 'size', which is independent of its direction. In the given exercise, the magnitude represents how far the vector extends in the plane. The magnitude is always a non-negative value and is calculated using the Pythagorean theorem for vectors. Specifically, for a vector \textbf{v} with components \( v_x \) and \( v_y \), the magnitude \( \|v\| \) is estimated using the formula:
\[ \|v\| = \sqrt{{v_x}^2 + {v_y}^2} \]
In practical terms, imagine walking in a straight line; the distance you've covered is analogous to the vector's magnitude. When the vector is in Cartesian form with given components, applying this formula will result in the vector's magnitude, which was originally known in the problem as 3. This concept is essential as it essentially converts the vector's characteristics into a scalar quantity, illustrating how much of the vector quantity there is.

Vector Components

Any vector in a plane can be represented by two components: one in the horizontal direction (\( v_x \), often associated with the \( i \) unit vector) and another in the vertical direction (\( v_y \), related to the \( j \) unit vector). These components are projections of the vector along the axes of a Cartesian coordinate system. In the provided exercise,
\[ v_x = 3\cos(60^{\bullet}) \]
\[ v_y = 3\sin(60^{\bullet}) \]
This shows that the vector \textbf{v} can be decomposed into a horizontal component of 1.5 and a vertical component of approximately 2.6. Understanding vector components is vital for tasks such as adding vectors, since it allows for the independent manipulation of each direction's influence.

Cartesian Form

When a vector is represented in terms of its horizontal (\( x \)) and vertical (\( y \)) components, it is said to be in 'Cartesian form'. This form allows us to easily visualize the vector in a traditional coordinate system, where the \( x \)-component and \( y \)-component correspond to the distances along the \( x \)-axis and \( y \)-axis, respectively. For the vector from the exercise, the Cartesian form is derived from its polar representation and is given by:
\[ \textbf{v} = v_x\textbf{i} + v_y\textbf{j} \]
This form makes it very straightforward to perform operation such as vector addition, finding scalar products, and more, because it aligns with the way we typically define coordinates and shapes within a plane.

Polar Form

The polar form of a vector presents it in terms of magnitude and direction angle, usually denoted \( r \) and \( \theta \), respectively. The magnitude (\( r \)) corresponds to the vector's length, while the direction angle (\( \theta \)) refers to the angle the vector makes with the positive \( x \)-axis. Converting the provided vector from polar to Cartesian form was a key solution step, since it involved using the cosine and sine of the direction angle alongside the magnitude to find the vector's components. Particularly, it uses the format:
\[ r(\cos \theta \textbf{i} + \sin \theta \textbf{j}) \]
The polar form is especially handy when dealing with problems related to angles and rotations, as it directly incorporates the angular position in its representation.