Problem 34 - University Calculus: Early Transcendentals

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

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

Step 1: Find the magnitude of the given vector

The vector given is \(\mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\). The magnitude of \(\mathbf{v}\) is calculated using the formula: \(\|\mathbf{v}\| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2}\). Evaluate the squares and sum them: \(\|\mathbf{v}\| = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}}\). This simplifies to \(\|\mathbf{v}\| = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).

Step 2: Normalize the given vector

To find a unit vector in the direction of \(\mathbf{v}\), divide each component of \(\mathbf{v}\) by its magnitude. The unit vector \(\hat{\mathbf{v}}\) is \(\hat{\mathbf{v}} = \frac{1}{\frac{\sqrt{3}}{2}} \left(\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\right)\). Simplifying, we get \(\hat{\mathbf{v}} = \frac{2}{\sqrt{3}} \left(\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\right)\) = \(\frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k}\).

Step 3: Find the vector in the opposite direction

To find the vector in the opposite direction, all signs of the unit vector's components are reversed. Therefore, the vector in the opposite direction is \(-\hat{\mathbf{v}} = -\frac{1}{\sqrt{3}}\mathbf{i} + \frac{1}{\sqrt{3}}\mathbf{j} + \frac{1}{\sqrt{3}}\mathbf{k}\).

Step 4: Scale the opposite vector to the desired magnitude

Multiply the unit vector in the opposite direction, \(-\hat{\mathbf{v}}\), by 3 (the desired magnitude). The resulting vector is \(3 \times (-\hat{\mathbf{v}}) = 3 \left(-\frac{1}{\sqrt{3}}\mathbf{i} + \frac{1}{\sqrt{3}}\mathbf{j} + \frac{1}{\sqrt{3}}\mathbf{k}\right)\). Simplifying gives us \(-\sqrt{3}\mathbf{i} + \sqrt{3}\mathbf{j} + \sqrt{3}\mathbf{k}\).

Key Concepts

Magnitude of a Vector

Finding the magnitude of a vector is one of the fundamental concepts in vector operations. Let's consider the vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \). To determine its magnitude, we use a formula that's reminiscent of the Pythagorean theorem:

• Formula: \( \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \)
• Apply: Substitute the components of our vector into the formula: \( \|\mathbf{v}\| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2} \).
• Simplify: This becomes \( \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).

The magnitude tells us how long the vector is, without regard to its direction. Understanding this allows us to further manipulate or understand the vector in different ways.

Unit Vector

A unit vector is a vector with a magnitude of 1, pointing in a particular direction. It's especially useful in defining direction alone, without magnitude scaling. To transform our vector \( \mathbf{v} \) into a unit vector, we divide each component of \( \mathbf{v} \) by its magnitude.

• Normalize: For \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \), the magnitude is \( \frac{\sqrt{3}}{2} \).
• Calculation: The unit vector \( \hat{\mathbf{v}} \) is computed as \( \hat{\mathbf{v}} = \frac{1}{\frac{\sqrt{3}}{2}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \right) \).
• Result: Simplifying this, we get \( \hat{\mathbf{v}} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \).

Unit vectors are critical building blocks in vector analysis, providing a basis for scaling and directionality.

Opposite Direction Vector

Vectors can point in many directions, but in some cases, you might want a vector that points exactly the opposite way, like a reversal. To find a vector in the opposite direction, we change the sign of each component.

• From Unit Vector: Start with the unit vector \( \hat{\mathbf{v}} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \).
• Opposite: Reversing the signs gives \( -\hat{\mathbf{v}} = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \).

This vector is the same length (1, since it's a unit vector) but points in the exact opposite direction. It's an indispensable concept when you want to analyze or define directional changes in vectors.

Scaling Vectors

Scaling vectors involves changing their magnitude while retaining their direction (or reversing it in case of negative scale factors). It's a powerful operation used to resize a vector easily.

• Desired Magnitude: In this exercise, we needed a vector of magnitude 3.
• Calculation: Multiply each component of \( -\hat{\mathbf{v}} \) by 3: \( 3 \left(-\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \right) \).
• Result: This gives the scaled vector \( -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \).

Scaling enables us to tailor vectors to any required size, which is crucial when those vectors need a specific impact or reach within a mathematical or physical analysis.