Problem 52 - Advanced Engineering Mathematics

Question

Find a vector \(\mathbf{b}\) for which \(\|\mathbf{b}\|=\frac{1}{2}\) that is parallel to \(\mathbf{a}=\langle-6,3,-2\rangle\) but has the opposite direction.

Step-by-Step Solution

Answer

The vector \( \mathbf{b} = \left( \frac{3}{7}, -\frac{3}{14}, \frac{1}{7} \right) \) has the desired properties.

Step 1: Understand the Problem

We need to find a vector \( \mathbf{b} \) that is parallel to \( \mathbf{a} = \langle -6, 3, -2 \rangle \), has a norm of \( \frac{1}{2} \), and points in the opposite direction of \( \mathbf{a} \).

Step 2: Calculate the Norm of Vector a

To start, find the norm (magnitude) of the vector \( \mathbf{a} = \langle -6, 3, -2 \rangle \). Use the formula \( \| \mathbf{a} \| = \sqrt{(-6)^2 + 3^2 + (-2)^2} \).

Step 3: Simplify the Norm

Calculate the expression: \( \| \mathbf{a} \| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \). Thus, the norm of \( \mathbf{a} \) is 7.

Step 4: Determine the Unit Vector in Opposite Direction

A unit vector in the opposite direction of \( \mathbf{a} \) is found by dividing \(-\mathbf{a}\) by \( \| \mathbf{a} \| \). Thus, \( \mathbf{u} = \left( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \right) \).

Step 5: Scale the Unit Vector

To find a vector with a norm of \( \frac{1}{2} \) in the direction of \( \mathbf{u} \), scale \( \mathbf{u} \) by \( \frac{1}{2} \). This gives us \( \mathbf{b} = \frac{1}{2} \mathbf{u} = \left( \frac{3}{7}, -\frac{3}{14}, \frac{1}{7} \right) \).

Step 6: Verify the Magnitude

Calculate \( \| \mathbf{b} \| \) to ensure it is \( \frac{1}{2} \). This involves checking \( \| \mathbf{b} \| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(-\frac{3}{14}\right)^2 + \left(\frac{1}{7}\right)^2} \), which is indeed \( \frac{1}{2} \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector, often denoted as \( \|\mathbf{v}\| \), is like the length of a line segment. It gives us a measure of how long the vector is. Think of it as the distance from the origin to the endpoint of the vector in a coordinate space. To find the magnitude, you can use the formula:

• For a 3D vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the magnitude is calculated as \( \sqrt{v_1^2 + v_2^2 + v_3^2} \).

In practice, if you have a vector \( \mathbf{a} = \langle -6, 3, -2 \rangle \), here's how you calculate its magnitude:

• Step 1: Square each component: \( (-6)^2 = 36 \), \( 3^2 = 9 \), \( (-2)^2 = 4 \).
• Step 2: Sum them up: \( 36 + 9 + 4 = 49 \).
• Step 3: Take the square root: \( \sqrt{49} = 7 \).

The result tells us that the vector \( \mathbf{a} \) has a magnitude of 7.

Unit Vector

A unit vector is a direction indicator, essentially showing the way a vector points without considering how long it is. It always has a magnitude of 1. This means no matter how large the original vector is, the unit vector is a smaller version pointing the same way.To get a unit vector \( \mathbf{u} \) in the direction of a given vector \( \mathbf{a} \), you divide each component of the vector \( \mathbf{a} \) by its magnitude. Consider our vector \( \mathbf{a} = \langle -6, 3, -2 \rangle \) with a magnitude of 7:

• To find the unit vector in the opposite direction, first take the negative of each component: \( \langle 6, -3, 2 \rangle \).
• Then, divide each component by 7 (its magnitude):
  • \( \frac{6}{7} \)
  • \( -\frac{3}{7} \)
  • \( \frac{2}{7} \)

Thus, the unit vector \( \mathbf{u} \) in the opposite direction of \( \mathbf{a} \) is \( \left( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \right) \). This unit vector retains the vector's direction but shrinks its scale to just 1.

Parallel Vectors

Parallel vectors are simply vectors that have the same or exactly opposite direction. They can be of any length, but they all point along the same line of direction.When two vectors are parallel, it means one vector is a scalar multiple of the other. This means you can multiply one vector by some number and get the second vector. For example, for vector \( \mathbf{b} \) to be parallel to \( \mathbf{a} \) and have the opposite direction, \( \mathbf{b} \) must be a negative scalar multiple of \( \mathbf{a} \).Let’s go through that with our specific example:

• We started with vector \( \mathbf{a} = \langle -6, 3, -2 \rangle \).
• The problem asks for a vector \( \mathbf{b} \) with the opposite direction, implying a need for a negative.
• First, convert \( \mathbf{a} \) to its unit vector in the opposite direction, as explained previously.
• Then, scale the unit vector to have a desired magnitude, in this case, \( \frac{1}{2} \).
• This is done by multiplying each component of the unit vector by \( \frac{1}{2} \).
  • This yields \( \mathbf{b} = \left( \frac{3}{7}, -\frac{3}{14}, \frac{1}{7} \right) \).

This whole process ensures \( \mathbf{b} \) is exactly parallel, oppositely directed, and correctly scaled compared to \( \mathbf{a} \).