Problem 20
Question
Express the given vector in terms of the unit vectors i, j, and k. $$ \langle 0,-3,5\rangle $$
Step-by-Step Solution
Verified Answer
-3j + 5k
1Step 1: Identify Components
The given vector is \( \langle 0, -3, 5 \rangle \). This means it has three components: the x-component is 0, the y-component is -3, and the z-component is 5.
2Step 2: Express in Terms of Unit Vectors
A vector \( \langle a, b, c \rangle \) can be expressed in terms of the unit vectors \( \mathbf{i, j, k} \) as \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \). For our vector, this becomes \( 0\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \).
3Step 3: Simplify Expression
Since multiplying a vector by zero results in a zero vector, \( 0\mathbf{i} \) can be omitted. Therefore, the vector expression simplifies to \( -3\mathbf{j} + 5\mathbf{k} \).
Key Concepts
Unit VectorsComponents of a VectorSimplifying Vectors
Unit Vectors
Unit vectors are the building blocks of vector notation. They are vectors of length one that point along the axes of a coordinate system. In three-dimensional space, we usually denote unit vectors as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
This approach makes it easier to handle and visualize vectors, as each component of a vector aligns with one of these principal directions.
- \( \mathbf{i} \) is the unit vector along the x-axis.
- \( \mathbf{j} \) is the unit vector along the y-axis.
- \( \mathbf{k} \) is the unit vector along the z-axis.
This approach makes it easier to handle and visualize vectors, as each component of a vector aligns with one of these principal directions.
Components of a Vector
Each vector in a 3D space can be broken down into its components along the x, y, and z axes. To understand this better, consider the vector \( \langle 0, -3, 5 \rangle \) from our exercise.
When expressed in terms of unit vectors, each component gets multiplied by its respective unit vector. This converts the components into a more practical and readable form, using \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), where \( a, b, \) and \( c \) are the components of the vector.
- x-component: This is 0, indicating that along the x-axis, there's no movement.
- y-component: This is -3, suggesting the vector moves 3 units in the negative y direction.
- z-component: This is 5, showing that it extends 5 units in the positive z direction.
When expressed in terms of unit vectors, each component gets multiplied by its respective unit vector. This converts the components into a more practical and readable form, using \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), where \( a, b, \) and \( c \) are the components of the vector.
Simplifying Vectors
Simplifying vectors is often necessary to make expressions cleaner and more readable.
Sometimes, the original expression of a vector may contain unnecessary terms, like multiplying a unit vector by zero. Consider the vector \( 0\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \). The term \( 0\mathbf{i} \) contributes nothing since multiplying by zero results in zero.To simplify:
Sometimes, the original expression of a vector may contain unnecessary terms, like multiplying a unit vector by zero. Consider the vector \( 0\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \). The term \( 0\mathbf{i} \) contributes nothing since multiplying by zero results in zero.To simplify:
- Identify terms: Look for any terms multiplied by zero, as they can be eliminated.
- Eliminate zero terms: In this case, remove \( 0\mathbf{i} \) because it adds no value.
- Rewrite the vector: After removing unnecessary terms, you get \( -3\mathbf{j} + 5\mathbf{k} \), which is a much cleaner expression of the vector.
Other exercises in this chapter
Problem 19
\(19-22\) . Sketch the given vector with initial point \((4,3),\) and find the terminal point. $$ \mathbf{u}=\langle 2,4\rangle $$
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A plane has normal vector \(\mathbf{n}\) and passes through the point \(P\) (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of th
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Determine whether the given vectors are perpendicular. $$ \mathbf{u}=4 \mathbf{i}, \quad \mathbf{v}=-\mathbf{i}+3 \mathbf{j} $$
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Describe the trace of the sphere $$ x^{2}+(y-4)^{2}+(z-3)^{2}=144 $$ in (a) the \(x\) -plane and in (b) the plane \(z=-2\)
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