Problem 25 - Advanced Engineering Mathematics

Question

Find a unit vector (a) in the same direction as \(\mathbf{a}\), and (b) in the opposite direction of \(\mathbf{a}\). \(\mathbf{a}=\langle 2,2\rangle\)

Step-by-Step Solution

Answer

The unit vector in the same direction is \(\left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\), and in the opposite direction, it is \(\left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle\).

Step 1: Determine Magnitude of Vector a

To find the unit vector in the same direction, we first need to find the magnitude of \(\mathbf{a}\). The formula for the magnitude of a vector \(\mathbf{a} = \langle x, y \rangle\) is \(\sqrt{x^2 + y^2}\). For \(\mathbf{a} = \langle 2, 2 \rangle\), the magnitude is \(\sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\).

Step 2: Find Unit Vector in the Same Direction

The unit vector \(\mathbf{u}\) in the same direction as \(\mathbf{a}\) is found by dividing each component of \(\mathbf{a}\) by its magnitude. Thus, \(\mathbf{u} = \left\langle \frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}} \right\rangle = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle\).

Step 3: Simplify the Unit Vector in the Same Direction

To simplify, we multiply the vector by \(\frac{\sqrt{2}}{\sqrt{2}}\) to rationalize the denominator: \(\left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\).

Step 4: Find Unit Vector in the Opposite Direction

To find the unit vector in the opposite direction, we simply negate the unit vector in the same direction. Thus, \(-\mathbf{u} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle\).

Key Concepts

Vector Magnitude

Understanding vector magnitude is crucial because it tells us about the size or length of a vector without considering its direction. Imagine a vector \(\mathbf{a} = \langle x, y \rangle\). The magnitude, often written as \(|\mathbf{a}|\) or \(||\mathbf{a}||\), can be thought of as the distance from the origin to the point \(\langle x, y \rangle\) in a coordinate system.
To calculate the magnitude, we employ the formula \(\sqrt{x^2 + y^2}\). Break it down like this:

• Square each component of the vector.
• Add these squared values.
• Take the square root of the sum.

For example, the magnitude of the vector \(\mathbf{a} = \langle 2, 2 \rangle\) is \(\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\). This tells you how long or how much ground the vector covers, regardless of where it points.

Vector Direction

The direction of a vector is the orientation or angle at which the vector points, distinguishing it from others. It's like telling a compass direction along a path. When you find a unit vector, you maintain the direction. However, you scale the vector to have a magnitude of 1. This process involves two simple steps:

• First, calculate the vector's magnitude.
• Next, divide each of the vector's components by this magnitude.

For our vector \(\mathbf{a} = \langle 2, 2 \rangle\), we found earlier the magnitude is \(2\sqrt{2}\). Thus, the unit vector becomes \(\left\langle \frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}} \right\rangle = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle\). We rationalize this to \(\left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\). This new vector keeps the direction of \(\mathbf{a}\) while having a length of 1. This is particularly useful in many fields of study, such as physics and engineering.

Opposite Vectors

Opposite vectors are straightforward yet powerful concepts. If you have a vector and imagine reversing its path entirely, you get its opposite vector. It's like retracing your steps exactly back to the starting point from where you are.
This is essential when calculating a vector in the opposite direction. Simply take the existing unit vector and multiply each of its components by \(-1\).
In our example, the unit vector in the direction of \(\mathbf{a}\) is \(\left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\). Its opposite would then be \(\left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle\).

• This opposite vector has the same length but points exactly in the reverse direction.
• Both vectors have the same magnitude, but their orientations differ by \(180^\circ\).

Knowing about opposite vectors proves quite handy when dealing with problems involving direction reversal, such as forces acting in opposite ways or travel routes in navigation.