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Magazine Chapter 11: Problem 12
Find two unit vectors that are parallel to the \(y z\) -plane and are orthogonal to the vector \(3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}\).
Short Answer
Expert verified
The unit vectors are \(\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)\) and \(\left(0, \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)\).
Step by step solution
01
Identifying the Properties
We need to find vectors that are both parallel to the \(yz\)-plane and orthogonal to \(3 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\). A vector in the \(yz\)-plane has no \(x\) component, so our vector will be of the form \(0\mathbf{i} + b\mathbf{j} + c\mathbf{k}\). The vector should also be orthogonal to \(3 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\), which means their dot product must be zero.
02
Setting Up the Dot Product Equation
The dot product of \(0\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) and \(3 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\) is \(0 \cdot 3 + b(-1) + c(2) = -b + 2c\). For orthogonality, we set this equal to zero: \[-b + 2c = 0.\]
03
Solving the Equation for Variables
From the equation \(-b + 2c = 0\), we can express \(b\) in terms of \(c\): \(b = 2c\). To simplify, let's choose \(c = 1\), then \(b = 2 \times 1 = 2\). The vector before normalization is \(0\mathbf{i} + 2\mathbf{j} + 1\mathbf{k}\).
04
Normalizing the Vector
The magnitude of the vector \(0\mathbf{i} + 2\mathbf{j} + 1\mathbf{k}\) is given by\[\sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}.\]To convert this into a unit vector, we divide each component by the magnitude:\[\mathbf{u}_1 = \left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right).\]
05
Finding the Orthogonal Pair
To find another unit vector orthogonal to \(3 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\) that is also in the \(yz\)-plane, we can choose \(b = 1\), then \(c = \frac{b}{2} = \frac{1}{2}\). The vector before normalization is \(0\mathbf{i} + 1\mathbf{j} + \frac{1}{2}\mathbf{k}\).Its magnitude is\[\sqrt{0^2 + 1^2 + (\frac{1}{2})^2} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.\]Normalize this vector:\[\mathbf{u}_2 = \left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right).\]
06
Normalizing the Second Vector
Divide each component of the vector \(0\mathbf{i} + 1\mathbf{j} + \frac{1}{2}\mathbf{k}\) by its magnitude \(\frac{\sqrt{5}}{2}\):\[\mathbf{u}_2 = \left(0, \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right).\]
07
Conclusion: Listing the Unit Vectors
The two unit vectors that are parallel to the \(yz\)-plane and orthogonal to \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) are:\[\mathbf{u}_1 = \left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right) \text{ and } \mathbf{u}_2 = \left(0, \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the yz-plane
The concept of the "yz-plane" in three-dimensional geometry is crucial for this problem. It refers to the plane where the x-coordinate is zero. In other words, any point or vector that lies completely on the yz-plane has no x-component. This characteristic simplifies many calculations because any vector in this plane can be represented as:
- Zero x-component: The vector is described as \(0\mathbf{i} + b\mathbf{j} + c\mathbf{k}\). The absence of the \(\mathbf{i}\) component indicates it's parallel to the yz-plane.
- Coordinates: The coordinates only involve y and z, making it (0, y, z).
Orthogonal Vectors Simplified
Two vectors are orthogonal when their dot product is zero. This relationship is important for vector calculations.
- Dot Product Rule: If you have vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), they are orthogonal when:\[a_1b_1 + a_2b_2 + a_3b_3 = 0\]
- Practical Use: For the exercises requiring orthogonality to a vector such as \(3 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\), it helps find suitable values for unknown components in the required vector.
Normalization Made Easy
In vector mathematics, normalization refers to the process of converting a vector into a unit vector. A unit vector has a magnitude of one, which can be useful in many mathematical contexts.
- Magnitude Calculation: First, find the magnitude of the vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\):\[\sqrt{a^2 + b^2 + c^2}\]
- Divide by Magnitude: To normalize, divide each component of the vector by this magnitude:\[\mathbf{u} = \left(\frac{a}{\text{magnitude}}, \frac{b}{\text{magnitude}}, \frac{c}{\text{magnitude}}\right)\]
- Practical Step: After calculation, the new coordinates give the unit vector, ensuring that its magnitude is precisely one.
