Chapter 11: Problem 34
Find a vector of magnitude 3 in the direction opposite to the direction of \(\mathbf{v}=(1 / 2) \mathbf{i}-(1 / 2) \mathbf{j}-(1 / 2) \mathbf{k}\)
Short Answer
Expert verified
The vector is \(-\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k}\).
Step by step solution
01
Find the Magnitude of the Given Vector
To find the magnitude of the vector \( \mathbf{v} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \), we can use the formula for the magnitude of a vector: \[ ||\mathbf{v}|| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 } \]This simplifies to:\[ ||\mathbf{v}|| = \sqrt{ \frac{1}{4} + \frac{1}{4} + \frac{1}{4} } = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \].
02
Normalize the Given Vector
To create a unit vector in the direction opposite to \( \mathbf{v} \), we first find the unit vector in the direction of \( \mathbf{v} \) by dividing each component by its magnitude. Thus, the unit vector \( \mathbf{u} \) is:\[ \mathbf{u} = \frac{1}{||\mathbf{v}||} (\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}) = \frac{2}{\sqrt{3}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \right)\]This simplifies to:\[ \mathbf{u} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \].
03
Reverse the Direction of the Unit Vector
Since we need a vector in the opposite direction, multiply the unit vector by -1:\[ -\mathbf{u} = -\left( \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \right) = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \].
04
Scale the Reversed Unit Vector to Desired Magnitude
Finally, to get a vector of magnitude 3, scale the reversed unit vector by 3:\[ \mathbf{w} = 3(-\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}) \]This results in:\[ \mathbf{w} = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
In vector mathematics, the magnitude of a vector, often referred to as its "length," is a measure of how long the vector is. This concept is crucial for understanding the size of vectors in physical space. To find the magnitude of a vector defined as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), you use the formula:
- \( ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \)
- \( ||\mathbf{v}|| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 } = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \)
Unit Vector
A unit vector is a vector that has a magnitude of 1 and points in a specific direction. It's like having a "standard" or "template" direction vector. Any vector can be converted into a unit vector by dividing each of its components by its magnitude. This process is known as normalization.
- Formula: \( \mathbf{u} = \frac{1}{||\mathbf{v}||} \cdot \mathbf{v} \)
- \( \mathbf{u} = \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \)
Direction Reversal
Reversing a vector's direction involves changing its orientation 180 degrees on its path. Mathematically, this is achieved by multiplying the vector by \(-1\). For instance, if \( \mathbf{u} \) is our unit vector, reversing its direction gives us:
- \( -\mathbf{u} = -\left( \frac{1}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{3}} \mathbf{k} \right) = -\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \)
Vector Scaling
Vector scaling involves changing the magnitude of a vector while preserving its direction. This can be achieved by multiplying the vector by a scalar value. For example, if we want a vector of magnitude 3 in the reversed direction of our unit vector \(-\mathbf{u}\), we simply multiply it by 3:
- \( \mathbf{w} = 3(-\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}) \)
- \( \mathbf{w} = -\sqrt{3} \mathbf{i} + \sqrt{3} \mathbf{j} + \sqrt{3} \mathbf{k} \)