Problem 34
Question
Double cancellation If \(\mathbf{u} \neq \mathbf{0}\) and if \(\mathbf{u} \times \mathbf{v}=\mathbf{u} \times \mathbf{w}\) and \(\mathbf{u} \cdot \mathbf{v}=\mathbf{u} \cdot \mathbf{w},\) then does \(\mathbf{v}=\mathbf{w} ?\) Give reasons for your answer.
Step-by-Step Solution
Verified Answer
Yes, \( \mathbf{v} = \mathbf{w} \); both conditions imply \( \mathbf{v} - \mathbf{w} = \mathbf{0} \).
1Step 1: Understand the Given Condition
We are provided with two vector conditions: 1. A cross product condition: \( \mathbf{u} \times \mathbf{v} = \mathbf{u} \times \mathbf{w} \) 2. A dot product condition: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} \)The vector \( \mathbf{u} \) is not the zero vector. Our goal is to determine if these conditions imply that \( \mathbf{v} = \mathbf{w} \).
2Step 2: Analyze the Cross Product Condition
The cross product condition \( \mathbf{u} \times \mathbf{v} = \mathbf{u} \times \mathbf{w} \) implies that the vectors \( \mathbf{v} - \mathbf{w} \) must be parallel to \( \mathbf{u} \). This is because the cross product of two vectors equals zero when they are scalar multiples of each other, indicating collinearity.
3Step 3: Analyze the Dot Product Condition
The dot product condition \( \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{w} \) translates to \( \mathbf{u} \cdot (\mathbf{v} - \mathbf{w}) = 0 \). This equation implies that the vector \( \mathbf{v} - \mathbf{w} \) is orthogonal to \( \mathbf{u} \).
4Step 4: Examine Both Conditions
For the vector \( \mathbf{v} - \mathbf{w} \) to satisfy both conditions - being parallel to and orthogonal to \( \mathbf{u} \), it can only be the zero vector. If \( \mathbf{v} - \mathbf{w} = \mathbf{0} \), then \( \mathbf{v} = \mathbf{w} \).
5Step 5: Conclusion
Given \( \mathbf{u} eq \mathbf{0} \), both conditions together imply that \( \mathbf{v} - \mathbf{w} \) must be the zero vector for it to be both parallel and orthogonal to \( \mathbf{u} \). Therefore, \( \mathbf{v} = \mathbf{w} \).
Key Concepts
Cross ProductDot ProductZero VectorParallel Vectors
Cross Product
The cross product is a mathematical operation that takes two vectors and returns another vector that is perpendicular to both of them. It is represented as \( \mathbf{u} \times \mathbf{v} \) and is widely used in physics and engineering to determine the vector orthogonal to a plane.
- The result of the cross product is a vector, not a scalar, and has a direction determined by the right-hand rule.
- The magnitude of the cross product reflects the area of the parallelogram formed by the two vectors.
Dot Product
The dot product, also known as the scalar product, involves two vectors and results in a scalar quantity. It is expressed as \( \mathbf{u} \cdot \mathbf{v} \) and is used to project one vector onto another.
- The dot product gives insight into how much one vector "goes in the direction" of another.
- A dot product of zero implies that the vectors are orthogonal, or at right angles to each other.
Zero Vector
The zero vector, often denoted as \( \mathbf{0} \), is a unique vector that has zero magnitude and is directionless. It serves as an additive identity in vector algebra, meaning any vector added to the zero vector remains unchanged.
- The zero vector is crucial in vector mathematics, symbolizing the concept of no movement or change in both position and magnitude.
Parallel Vectors
Parallel vectors are vectors that have the same or exact opposite direction. They can be any magnitude but must maintain the same direction. When vectors are parallel, they are scalar multiples of each other.
- If \( \mathbf{a} \) and \( \mathbf{b} \) are parallel, then there exists a scalar \( k \) such that \( \mathbf{a} = k \mathbf{b} \).
Other exercises in this chapter
Problem 33
Write inequalities to describe the sets in Exercises \(29-34\) The (a) interior and (b) exterior of the sphere of radius 1 centered at the point \((1,1,1)\)
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Line parallel to a vector Show that the vector \(\mathbf{v}=a \mathbf{i}+b \mathbf{j}\) is parallel to the line \(b x-a y=c\) by establishing that the slope of
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In Exercises 33–38, find the distance from the point to the line. $$ (0,0,0) ; \quad x=5+3 t, \quad y=5+4 t, \quad z=-3-5 t $$
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