Problem 23 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=\sqrt{10}\); when drawn in standard position \(\vec{v}\) lies in Quadrant II and makes an angle measuring arctan(3) with the negative \(x\) -axis

Step-by-Step Solution

Answer

The component form of the vector is \((-1, 3)\).

Step 1: Understand the vector's quadrant and angle

The vector \(\vec{v}\) lies in Quadrant II. Since it makes an angle of \(\arctan(3)\) with the negative x-axis, this means the vector's direction is towards the positive y-axis as angles in Quadrant II are measured counterclockwise from the negative x-axis.

Step 2: Determine angle from positive x-axis

If the vector makes an angle \(\arctan(3)\) with the negative x-axis, then the angle from the positive x-axis can be expressed as \(\pi - \arctan(3)\) because a full half rotation from the positive x-axis to the negative x-axis is \(\pi\) radians.

Step 3: Express the trigonometric values

Given \(\theta = \pi - \arctan(3)\), calculate the cosine and sine. In Quadrant II, \(\cos(\theta) = -\cos(\arctan(3))\), and \(\sin(\theta) = \sin(\arctan(3))\), accounting for the direction signs (negative for cosine, positive for sine in Quadrant II).

Step 4: Use trigonometric identities

Given \(\tan(\theta) = 3\), we can set \(\tan(\theta) = \frac{3}{1}\). This forms a right triangle with opposite side 3, adjacent side 1, and hypotenuse \(\sqrt{10}\). Hence, \(\cos(\theta) = -\frac{1}{\sqrt{10}}, \sin(\theta) = \frac{3}{\sqrt{10}}\).

Step 5: Calculate vector components

Use the magnitude of \(\vec{v}\) which is \(\sqrt{10}\) and its component form formula: \(\|\vec{v}\| \cdot \cos(\theta)\) for the x-component, and \(\|\vec{v}\| \cdot \sin(\theta)\) for the y-component. Thus, the components are \(-\frac{1}{\sqrt{10}} \cdot \sqrt{10} = -1\) for x, and \(\frac{3}{\sqrt{10}} \cdot \sqrt{10} = 3\) for y.

Step 6: Write the vector in component form

The component form of vector \(\vec{v}\) is \((-1, 3)\).

Key Concepts

Magnitude and Direction of a Vector

Understanding vectors means knowing both their magnitude and direction. The magnitude of a vector is like its length or size. We measure it using the Pythagorean theorem. For example, if vector \(\vec{v}\) has a magnitude of \(\sqrt{10}\), it means its length is like having a hypotenuse of a triangle with sides that square to make 10.

The direction of a vector tells us where it's pointing. This is where angles come in. When vectors are in the Cartesian plane, the angle direction helps us understand which Quadrant the vector lies in. Knowing both these facts helps us write vectors easily in their component forms, which is a bit like finding their address. This specifies exactly where the vector points along the x and y axes.

Trigonometric Identities

Trigonometric identities are like puzzle pieces, showing relationships between functions like sine, cosine, and tangent. These identities are the key tools we use to find angles and sides in triangles, especially when dealing with vectors.

For example, if you know \(\tan(\theta) = 3\), it forms part of the clue in a right triangle's opposite and adjacent sides. Once these sides are known, you can find \(\sin(\theta)\) and \(\cos(\theta)\). In our exercise, the vector \(\vec{v}\) uses these identities to determine its components on the x and y-axis. Here's where cosine becomes the horizontal part and sine the vertical. Understanding these trigonometric identities leads to precise calculations and an accurate component form of vectors.

Arctan Function

The arctan function is critical when working backward from a tangent to find an angle. It's called the 'inverse tangent' and helps to determine which angle has a specific tangent ratio.

When the problem states that there's a vector with a direction of \(\arctan(3)\), it means you can reverse the process of finding tangent. You'll find which angle results in a tangent of 3. This angle helps decide the vector's position relative to the x-axis. In our case, solving for \(\arctan(3)\) gives a precise angle that helps calculate the vector's direction in the correct quadrant.

Quadrants in Trigonometry

Quadrants are the divisions of the Cartesian plane that help us know where an angle or vector stands. Typically, there are four quadrants:

• Quadrant I: both x and y are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: both x and y are negative.
• Quadrant IV: x is positive, y is negative.

This system helps determine the signs of sine, cosine, and tangent. For vectors, knowing the quadrant means you understand a lot about its direction. In the case of vector \(\vec{v}\) in Quadrant II, it alerts us to which trigonometric signs to use — cosine being negative while sine remains positive. This is crucial in finding the vector's component form correctly and highlights the logical path in deciding the vector's precise characteristics.