Problem 28 - Advanced Engineering Mathematics

Question

Find a unit vector (a) in the same direction as \(\mathbf{a}\), and (b) in the opposite direction of \(\mathbf{a}\). \(\mathbf{a}=\langle 1,-\sqrt{3}\rangle\)

Step-by-Step Solution

Answer

(a) In the same direction: \( \langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \); (b) Opposite direction: \( \langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \).

Step 1: Calculate the Magnitude of Vector

To find a unit vector in the direction of a given vector \( \mathbf{a} = \langle 1, -\sqrt{3} \rangle \), we first need to calculate the magnitude of \( \mathbf{a} \). The formula for the magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is given by:\[ ||\mathbf{v}|| = \sqrt{x^2 + y^2} \]Applying the formula to \( \mathbf{a} \):\[ ||\mathbf{a}|| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Step 2: Find Unit Vector in the Same Direction

A unit vector in the same direction as \( \mathbf{a} \) is obtained by dividing each component of \( \mathbf{a} \) by its magnitude:\[ \mathbf{u} = \frac{1}{||\mathbf{a}||} \mathbf{a} = \frac{1}{2} \langle 1, -\sqrt{3} \rangle \]Thus, we have:\[ \mathbf{u} = \langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \]

Step 3: Find Unit Vector in the Opposite Direction

To find a unit vector in the opposite direction of \( \mathbf{a} \), we take the unit vector in the same direction \( \mathbf{u} \) and multiply it by \(-1\):\[ \mathbf{v} = -\mathbf{u} = -\langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle = \langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \]

Key Concepts

Vector Magnitude

The magnitude of a vector is like the length of a line. For a vector given in a two-dimensional plane, such as \( \mathbf{a} = \langle 1, -\sqrt{3} \rangle \), the magnitude provides an idea of how long the vector is. To find this length, we use the formula \( ||\mathbf{v}|| = \sqrt{x^2 + y^2} \). Here, we are essentially applying the Pythagorean theorem to the vector components, treating them like the two sides of a right triangle.

For \( \mathbf{a} \), with components 1 and \(-\sqrt{3}\):

• First, square each component: \(1^2 = 1\) and \((-\sqrt{3})^2 = 3\).
• Next, add these squares: \(1 + 3 = 4\).
• Finally, take the square root of this sum: \(\sqrt{4} = 2\).

The magnitude from this calculation, \(||\mathbf{a}|| = 2\), tells us the vector’s total length, which is crucial for finding a unit vector.

Vector Direction

Once you have the magnitude, the direction of the vector is defined by its unit vector. A unit vector maintains the original direction but transforms the length to 1. This is done by dividing each component of the original vector by its magnitude.

For vector \( \mathbf{a} = \langle 1, -\sqrt{3} \rangle \), here’s how we find its unit vector:

• Divide the x-component (1) by the magnitude (2): \( \frac{1}{2} \).
• Divide the y-component \((-\sqrt{3})\) by the magnitude (2): \( -\frac{\sqrt{3}}{2} \).

This results in the unit vector \( \mathbf{u} = \langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \), which points in the same direction as \( \mathbf{a} \) but has a length of 1. A unit vector is essential for applications requiring a direction without concern for magnitude, such as defining directions in space or standardizing vectors.

Opposite Vector

Finding the opposite vector involves changing the direction to its reverse while keeping the length constant, when a unit vector is needed in the opposite direction. To get this vector, we take the unit vector already calculated and multiply each of its components by \(-1\). This flips the vector to point in exactly the opposite direction.

Starting with \( \mathbf{u} = \langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \rangle \), the steps are:

• Negate the x-component: \(-\frac{1}{2}\).
• Negate the y-component: \(\frac{\sqrt{3}}{2}\).

Thus, the opposite unit vector is \( \mathbf{v} = \langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \). This vector maintains the unit length but points exactly 180 degrees from the original direction of \( \mathbf{a} \).