Problem 16
Question
Find the vectors \(\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v},\) and \(3 \mathbf{u}-\frac{1}{2} \mathbf{v}\) $$\mathbf{u}=\langle 0,1,-3\rangle, \quad v=\langle 4,2,0\rangle$$
Step-by-Step Solution
Verified Answer
\(\mathbf{u}+\mathbf{v}=\langle 4,3,-3\rangle\), \(\mathbf{u}-\mathbf{v}=\langle -4,-1,-3\rangle\), \(3\mathbf{u}-\frac{1}{2}\mathbf{v}=\langle -2,2,-9\rangle\)."
1Step 1: Define the vectors
Start by defining the given vectors. We have \( \mathbf{u} = \langle 0, 1, -3 \rangle \) and \( \mathbf{v} = \langle 4, 2, 0 \rangle \).
2Step 2: Calculate \(\mathbf{u} + \mathbf{v}\)
We add the corresponding components of vectors \(\mathbf{u}\) and \(\mathbf{v}\). \[ \mathbf{u} + \mathbf{v} = \langle 0+4, 1+2, -3+0 \rangle = \langle 4, 3, -3 \rangle \]
3Step 3: Calculate \(\mathbf{u} - \mathbf{v}\)
Next, we subtract the corresponding components of \(\mathbf{v}\) from \(\mathbf{u}\).\[ \mathbf{u} - \mathbf{v} = \langle 0-4, 1-2, -3-0 \rangle = \langle -4, -1, -3 \rangle \]
4Step 4: Calculate \(3 \mathbf{u}\) and \(-\frac{1}{2} \mathbf{v}\) separately
First, calculate \(3 \mathbf{u}\) by multiplying each component of \(\mathbf{u}\) by 3.\[ 3 \mathbf{u} = \langle 3 \times 0, 3 \times 1, 3 \times (-3) \rangle = \langle 0, 3, -9 \rangle \]Then, calculate \(-\frac{1}{2} \mathbf{v}\) by multiplying each component of \(\mathbf{v}\) by \(-\frac{1}{2}\).\[ -\frac{1}{2} \mathbf{v} = \langle -\frac{1}{2} \times 4, -\frac{1}{2} \times 2, -\frac{1}{2} \times 0 \rangle = \langle -2, -1, 0 \rangle \]
5Step 5: Calculate \(3 \mathbf{u} - \frac{1}{2} \mathbf{v}\)
Finally, add the vectors obtained in Step 4.\[ 3 \mathbf{u} - \frac{1}{2} \mathbf{v} = \langle 0, 3, -9 \rangle + \langle -2, -1, 0 \rangle = \langle 0 + (-2), 3 + (-1), -9 + 0 \rangle = \langle -2, 2, -9 \rangle \]
Key Concepts
Vector SubtractionScalar MultiplicationThree-Dimensional Vectors
Vector Subtraction
Subtraction of vectors involves taking each component of one vector and subtracting the corresponding component of another vector. Imagine you have two vectors, for instance, \( \mathbf{u} = \langle 0, 1, -3 \rangle \) and \( \mathbf{v} = \langle 4, 2, 0 \rangle \). To find \( \mathbf{u} - \mathbf{v} \), you subtract each component of \( \mathbf{v} \) from each component of \( \mathbf{u} \) like so:
Vector subtraction is a simple yet powerful tool for finding differences between directions and magnitudes in space.
- First component: \( 0 - 4 = -4 \)
- Second component: \( 1 - 2 = -1 \)
- Third component: \( -3 - 0 = -3 \)
Vector subtraction is a simple yet powerful tool for finding differences between directions and magnitudes in space.
Scalar Multiplication
Scalar multiplication involves multiplying each component of a vector by a scalar (a constant number). Let's imagine vector \( \mathbf{u} = \langle 0, 1, -3 \rangle \) and we want to find \( 3 \mathbf{u} \), which means multiplying the vector by the scalar 3. Here's how it works:
- Multiply the first component by 3: \( 3 \times 0 = 0 \)
- Multiply the second component by 3: \( 3 \times 1 = 3 \)
- Multiply the third component by 3: \( 3 \times -3 = -9 \)
Three-Dimensional Vectors
Three-dimensional vectors are all about navigating through space in three directions: x, y, and z. Each component of a 3D vector represents a dimension in this space. Vectors like \( \mathbf{u} = \langle 0, 1, -3 \rangle \) live in a world where each of those numbers
represents movement along one of these axes.
Three-dimensional vectors give us the mathematical tools to represent real-world spatial relationships in calculations and visualizations.
represents movement along one of these axes.
- The x-component tells you how far right or left to go
- The y-component describes how far up or down
- The z-component, often the trickiest, indicates movement backward or forward out of our familiar two-dimensional view
Three-dimensional vectors give us the mathematical tools to represent real-world spatial relationships in calculations and visualizations.
Other exercises in this chapter
Problem 16
The lengths of two vectors a and \(b\) and the angle \(\theta\) between them are given. Find the length of their cross product, \(|\mathbf{a} \times \mathbf{b}|
View solution Problem 16
A plane has normal vector \(n\) and passes through the point \(P\). (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of the plane.
View solution Problem 16
Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}+4 x-6 y+2 z=10$$
View solution Problem 16
Determine whether the given vectors are perpendicular. $$\mathbf{u}=\langle 0,-5\rangle, \quad \mathbf{v}=\langle 4,0\rangle$$
View solution