Chapter 12: Problem 37
Let \(\mathbf{u}=\langle 3,-4\rangle, \mathbf{v}=\langle 1,1\rangle,\) and \(\mathbf{w}=\langle-1,0\rangle .\) Carry out the following computations. Find \(|2 \mathbf{u}+3 \mathbf{v}-4 \mathbf{w}|\)
Short Answer
Expert verified
Answer: The magnitude is \(\sqrt{194}\).
Step by step solution
01
Perform Vector Operations
First, we need to calculate \(2 \mathbf{u}+3 \mathbf{v}-4 \mathbf{w}\). To do this, we perform scalar multiplication and addition/subtraction as follows:
\(2 \mathbf{u} = 2\langle 3, -4 \rangle = \langle 6, -8 \rangle\)
\(3 \mathbf{v} = 3\langle 1, 1 \rangle = \langle 3, 3 \rangle\)
\(4 \mathbf{w} = 4\langle -1, 0 \rangle= \langle -4,0\rangle\)
Now, we perform the addition and subtraction:
\(2 \mathbf{u}+3 \mathbf{v}-4 \mathbf{w} = \langle 6, -8 \rangle + \langle 3, 3 \rangle - \langle -4, 0 \rangle\)
02
Add/Subtract the Vectors
Next, we combine the vectors component-wise:
\(\begin{aligned} &\langle 6, -8 \rangle + \langle 3, 3 \rangle - \langle -4, 0 \rangle \\ =& \langle 6 + 3 - (-4), -8 + 3 + 0 \rangle \\ =& \langle 13, -5 \rangle \end{aligned}\)
So, the resulting vector is \(\mathbf{r} = \langle 13, -5 \rangle\).
03
Calculate the Magnitude
Finally, we need to find the magnitude of the resulting vector \(\mathbf{r}\). The magnitude is determined using the formula \(| \mathbf{r} | = \sqrt{r_{x}^{2} + r_{y}^{2}}\), where \(r_{x}\) and \(r_{y}\) are the components of \(\mathbf{r}\).
\(| \mathbf{r} | = | \langle 13, -5 \rangle | = \sqrt{(13)^{2} + (-5)^{2}} = \sqrt{169 + 25} = \sqrt{194}\)
Therefore, the magnitude of \(|2 \mathbf{u}+3 \mathbf{v}-4 \mathbf{w}|\) is \(\sqrt{194}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
Understanding the magnitude of a vector is central to problem-solving in vector calculus. The magnitude, sometimes called the length or norm, represents the distance of the vector from the origin in a coordinate system. To compute it, use the formula:\[| \mathbf{r} | = \sqrt{r_{x}^{2} + r_{y}^{2}}\]This formula involves taking the square root of the sum of each component squared. For example, if we consider a vector \( \mathbf{r} = \langle 13, -5 \rangle \), its magnitude is calculated as:\[| \mathbf{r} | = \sqrt{(13)^{2} + (-5)^{2}} = \sqrt{194}\]Remember:
- The magnitude is always non-negative.
- It's a measure of length, so it doesn’t have any directional property.
- Although we only considered 2-D vectors here, this formula extends to 3-D and beyond by adding additional squared components.
Vector Addition
Vector addition involves combining vectors to yield a resultant vector by adding their corresponding components. It is a straightforward yet powerful tool in vector calculus.To add two vectors, such as \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), perform component-wise addition:\[\mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle\]For instance, in our exercise:
- First, we find \( 2 \mathbf{u} = \langle 6, -8 \rangle \)
- Next, \( 3 \mathbf{v} = \langle 3, 3 \rangle \)
- Final step involves combining these with \( -4 \mathbf{w} = \langle -4, 0 \rangle \)
- Vectors remain unaffected in their directional properties through addition.
- Always pay special attention to signs while summing components.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a real number). This operation affects the magnitude of the vector but not its direction unless the scalar is negative, which would reverse the direction.For a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), multiplying by a scalar \( c \) results in:\[c \mathbf{v} = \langle c\cdot v_1, c\cdot v_2 \rangle\]Specific transformations include:
- Stretching the vector when \( c > 1 \).
- Shrinking it if \( 0 < c < 1 \).
- Flipping its direction when \( c < 0 \).
Vector Operations
Vector operations include a variety of tasks such as addition, subtraction, and multiplication by scalars. These operations form the basis of vector algebra and help solve complex problems in physics and engineering.During operations:- **Addition** combines vectors, maintaining the overall direction while adjusting magnitude.- **Subtraction** involves adding a vector pointing in the opposite direction.- **Multiplication by scalars** involves scaling the vector without changing the direction, unless the scalar is negative.For instance, in the given example:
- We multiplied vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) by respective scalars to manipulate their magnitudes.
- Added and subtracted the results to streamline the combination.