Problem 34 - Thomas Calculus

Question

Find a vector of magnitude 3 in the direction opposite to the direction of \(\mathbf{v}=(1 / 2) \mathbf{i}-(1 / 2) \mathbf{j}-(1 / 2) \mathbf{k} .\)

Step-by-Step Solution

Answer

A vector of magnitude 3 in the opposite direction is \(-\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k}\).

Step 1: Determine the Unit Vector in the Direction of \(\mathbf{v}\)

The given vector is \(\mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\). First, find its magnitude: \(\|\mathbf{v}\| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). The unit vector in the direction of \(\mathbf{v}\) is \(\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \mathbf{i} - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \mathbf{j} - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \mathbf{k} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k}\).

Step 2: Reverse the Direction of the Unit Vector

To find the direction opposite to \(\mathbf{v}\), we reverse the unit vector \(\mathbf{u}\). This gives us \(-\mathbf{u} = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}\).

Step 3: Scale the Opposite Unit Vector to Have a Magnitude of 3

Multiply the reversed unit vector by the desired magnitude to find the final vector:\(\mathbf{w} = 3(-\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}) = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k}\).

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude (or length) of exactly 1 unit.
This means it simply points in the direction of a vector without stretching or shrinking it.
Because of its standardized magnitude, the unit vector is extremely useful in many vector calculations, such as finding components and direction reversals.To find a unit vector in the direction of any given vector, we divide each component of the vector by its magnitude.
For the vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \), the magnitude was calculated as \( \frac{\sqrt{3}}{2} \).
Hence, the unit vector is determined by: * Dividing each component of \( \mathbf{v} \) by \( \frac{\sqrt{3}}{2} \). * This results in the unit vector \( \mathbf{u} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \). With this unit vector, you accurately maintain the vector's direction while simplifying the calculations with other scalar values, like reversing direction or changing its magnitude.

Magnitude

The magnitude of a vector represents its length or size, similar to measuring how long something is.
In three-dimensional space, like with our vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \), the magnitude is found using the formula:\[ \|\mathbf{v}\| = \sqrt{ \, x^2 + y^2 + z^2} \] where \( x, y, \) and \( z \) are the components of the vector.
When calculating this formula for the given vector, we apply it to get\[ \| \mathbf{v} \| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Finding the magnitude helps in several ways:

• Determining the scale when making a unit vector.
• Establishing the size of a vector when scaled, such as when you need a vector of a specified length.
• Facilitating operations, like addition and subtraction of vectors, to obtain results that physically sense.

Knowing the magnitude allows us to effortlessly resize the vector, helping in activities like direction reversal or establishing vectors with desired properties.

Direction Reversal

The process of reversing the direction of a vector is straightforward yet powerful.
When you reverse a vector, you flip its direction entirely, akin to pointing in the exact opposite direction.This operation is accomplished by negating each component of the vector.
In the exercise, to find a vector in the opposite direction of the unit vector \( \mathbf{u} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \), we simply reverse each component to get:

• \( -\mathbf{u} = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \)

Implementing direction reversal is crucial when:

• You need to solve problems involving forces, motions, or velocities heading opposite the original path.
• Vectors must cancel each other out, like in determining equilibrium states in physics.

By reversing the direction and then rescaling it, you can construct vectors like \( \mathbf{w} \), which fulfills specific spatial and directional criteria. This ability completes a wide array of practical calculations in mathematics and scientific disciplines.