Problem 32 - Advanced Engineering Mathematics

Question

Find a vector \(b\) that is parallel to the given vector and has the indicated magnitude. \(\mathbf{a}=\frac{1}{2} \mathbf{i}-\frac{1}{2} \mathbf{j},\|\mathbf{b}\|=3\)

Step-by-Step Solution

Answer

\( \mathbf{b} = \frac{3\sqrt{2}}{4} \mathbf{i} - \frac{3\sqrt{2}}{4} \mathbf{j} \)

Step 1: Find the Magnitude of Vector a

The first step is to calculate the magnitude of vector \( \mathbf{a} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \). Use the formula for the magnitude of a vector: \( \| \mathbf{a} \| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 } = \sqrt{ \frac{1}{4} + \frac{1}{4} } = \sqrt{ \frac{1}{2} } = \frac{1}{\sqrt{2}} \).

Step 2: Find the Unit Vector in the Direction of a

To find a unit vector \( \mathbf{u} \) in the direction of \( \mathbf{a} \), divide \( \mathbf{a} \) by its magnitude: \( \mathbf{u} = \frac{\mathbf{a}}{\| \mathbf{a} \|} = \frac{1}{\sqrt{2}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \right) = \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \).

Step 3: Scale the Unit Vector to the Desired Magnitude

Now, multiply the unit vector \( \mathbf{u} \) by the desired magnitude of \( 3 \) to find \( \mathbf{b} \): \( \mathbf{b} = 3 \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - 3 \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} = \frac{3}{\sqrt{8}} \mathbf{i} - \frac{3}{\sqrt{8}} \mathbf{j} \).

Step 4: Simplify the Expression for b

Simplify \( \frac{3}{\sqrt{8}} \) by rationalizing the denominator: \( \frac{3}{\sqrt{8}} = \frac{3 \times \sqrt{8}}{8} = \frac{3\sqrt{8}}{8} = \frac{3\sqrt{2}\sqrt{4}}{8} = \frac{3\times 2\sqrt{2}}{8} = \frac{3\sqrt{2}}{4} \). Thus, vector \( \mathbf{b} = \frac{3\sqrt{2}}{4} \mathbf{i} - \frac{3\sqrt{2}}{4} \mathbf{j} \).

Key Concepts

Vector Magnitude

Understanding the magnitude of a vector is crucial in vector calculus. The magnitude, which is often referred to as the "length" or "norm" of a vector, is a representation of the vector's size or extent. For a vector represented in Cartesian coordinates as \( \mathbf{a} = x \mathbf{i} + y \mathbf{j} \), its magnitude can be calculated using the formula:

• \( \| \mathbf{a} \| = \sqrt{x^2 + y^2} \)

This formula is derived from the Pythagorean theorem, as one treats vectors in 2D like the hypotenuse of a right triangle. In the given example, vector \( \mathbf{a} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \), the magnitude was calculated as \( \frac{1}{\sqrt{2}} \). Using vector magnitude helps us understand how long a vector is, which becomes particularly useful when performing operations like normalization or scaling.

Unit Vector

A unit vector is a vector of length 1 that indicates direction. It is very useful in physics and engineering for representing direction without regard to magnitude. To convert any vector into a unit vector, you need to divide the vector by its magnitude.

• For vector \( \mathbf{a} \), the unit vector \( \mathbf{u} \) is given by: \( \mathbf{u} = \frac{\mathbf{a}}{\| \mathbf{a} \|} \)

In the solution steps, we made vector \( \mathbf{a} \) into a unit vector:

• \( \mathbf{u} = \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \)

This unit vector maintains the direction of original vector \( \mathbf{a} \), but with a magnitude of 1. This is crucial for various vector operations, such as scaling vectors to certain lengths.

Parallel Vectors

Vectors that have the same or exact opposite direction are considered parallel. Two vectors are parallel if one is a scalar multiple of the other. This means if you multiply one vector by a constant, you should be able to get the other.

• For example, if \( \mathbf{a} = k \mathbf{b} \), where \( k \) is any scalar, they are parallel.

In our solution, since vector \( \mathbf{b} \) was required to be parallel to \( \mathbf{a} \), we used the scalar multiple property. By scaling the unit vector derived from vector \( \mathbf{a} \) using a specific magnitude provided (3), we ensured \( \mathbf{b} \) points in the same direction as \( \mathbf{a} \).

Vector Scaling

Scaling a vector involves changing its magnitude without altering its direction. This operation is crucial when you wish to stretch or shrink a vector by a certain factor. To scale a unit vector to a specific magnitude, simply multiply it by the desired magnitude.

• If \( \mathbf{u} \) is a unit vector, to scale it to a magnitude \( k \), multiply: \( \mathbf{v} = k \mathbf{u} \)

For the problem at hand, the unit vector from \( \mathbf{a} \) was scaled to reach a magnitude of 3, resulting in:

• \( \mathbf{b} = 3\left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - 3\left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \)

Vector scaling helps in multiple applications, such as adjusting forces or velocities in physics simulations to meet certain criteria.