Question

Find the unit vector in the second quadrant of \(\mathbb{R}^{2}\) that makes a \(30^{\circ}\) angle with the \(y\) -axis.

Short answer

The unit vector is \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).

Step-by-step solution

Understanding the Problem

We need to find a unit vector in \(\mathbb{R}^2\) that lies in the second quadrant and forms a \(30^\circ\) angle with the \(y\)-axis. The second quadrant is where \(x < 0\) and \(y > 0\).

Determine the Angle from the Positive X-Axis

Since the angle with the \(y\)-axis is given, and we are in the second quadrant, convert the \(30^\circ\) angle from the \(y\)-axis. In the second quadrant, the angle from the positive x-axis would be \(180^\circ - 30^\circ = 150^\circ\).

Calculate the Components of the Unit Vector

A unit vector \(\mathbf{u}\) making an angle \(\theta\) with the positive x-axis is given by \(\mathbf{u} = (\cos\theta, \sin\theta)\). Thus, for \(\theta = 150^\circ\), we find: \(\cos 150^\circ = -\frac{\sqrt{3}}{2}\) and \(\sin 150^\circ = \frac{1}{2}\).

Write the Unit Vector

The unit vector \(\mathbf{u}\) in the second quadrant making a \(30^\circ\) angle with the \(y\)-axis is \(\mathbf{u} = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).

Verify the Length of the Unit Vector

Calculate the magnitude of \(\mathbf{u}\): \[\| \mathbf{u} \| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1.\] This confirms that \(\mathbf{u}\) is indeed a unit vector.

Key concepts

Second Quadrant

When working with trigonometry in \( ext{ }\mathbb{R}^2\text{ }\), it's essential to understand the four quadrants of the Cartesian plane. The second quadrant is a crucial location where certain vector directions and angles are defined. It is the area where the \( ext{ }x ext{ }\)-coordinates are negative, and the \( ext{ }y ext{ }\)-coordinates are positive.
This means any vector located here will point upward to the left, crossing over the y-axis but staying on the left side of the Cartesian plane.

This is important because it helps limit the possible angles and directions a vector can have, especially when given situations that involve specific angles. Understanding these constraints allows you to correctly determine vector directions and calculate associated angles.

Angle with Y-Axis

To find vectors related to specific angles, it's often preferable to look at reference angles first.
For instance, an angle given with respect to the \( ext{ }y ext{ }\)-axis might require a bit of readjustment when positioning a vector in the correct quadrant.

In our exercise, the angle provided is \(30^\circ\) concerning the \( ext{ }y ext{ }\)-axis, but to understand it in terms of positioning from the positive \( ext{ }x ext{ }\)-axis, we have to relate it to the familiar full rotation of \(180^\) degrees within the quadrant. This translates into an angle of \(180^\circ - 30^\circ = 150^\circ\) with respect to the positive \( ext{ }x ext{ }\)-axis.

This calculation ensures that the vector direction aligns with the constraints of lying in the second quadrant.

Magnitude of Vector

The magnitude, or length, of a vector is a fundamental property that defines how long the vector is in \( ext{ }\mathbb{R}^2\text{ }\). For unit vectors, the magnitude must always be \( ext{ }1\text{ }\).

To verify this attribute, you use the formula: \[\| \mathbf{u} \| = \sqrt{x^2 + y^2}\text{ }\].In our case, the components given were \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\). Calculating the magnitude, we check if these components indeed reflect a unit vector: \[\| \mathbf{u} \| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1\].

Ensuring the components are correct is key in confirming that the vector remains within the constraints of being a unit vector.

Trigonometry in Vectors

Understanding vectors often involves trigonometric concepts to determine their direction and position on a plane.
In particular, the relationship between the angles and the corresponding vector components—namely cosine and sine values—is essential.

For a unit vector forming an angle \((\theta)\) with the positive \( ext{ }x ext{ }\)-axis, it is expressed as \(\mathbf{u} = (\cos\theta, \sin\theta)\).
This means that for any angle, the cosine and sine functions determine the proportional x and y components of the vector. For example, at an angle of \(150^\circ\), the vector's components are \(\cos 150^\circ = -\frac{\sqrt{3}}{2}\) and \(\sin 150^\circ = \frac{1}{2}\).

Through these calculations, we pinpoint exactly the vector's placement and direction while maintaining the unit property. Thus, an understanding of trigonometry provides not just the complete vector representation but also reinforces its adequacy in specified positions and properties.