Problem 46
Question
Given \(\mathbf{u}=\langle-2,-3\rangle\) and \(\mathbf{v}=\langle-3,-5\rangle\) find \(3 \mathbf{u}-\mathbf{v}\).
Step-by-Step Solution
Verified Answer
The result is \(\langle -3, -4 \rangle\).
1Step 1: Multiply Vector \(\mathbf{u}\) by 3
Given the vector \(\mathbf{u} = \langle -2, -3 \rangle\), calculate \(3 \mathbf{u}\). Multiply each component of \(\mathbf{u}\) by 3: \[3 \mathbf{u} = 3 \times \langle -2, -3 \rangle = \langle 3 \times (-2), 3 \times (-3) \rangle = \langle -6, -9 \rangle\]
2Step 2: Write Down Vector \(\mathbf{v}\)
Identify vector \(\mathbf{v}\) as \(\langle -3, -5 \rangle\). This vector will be subtracted from the result of Step 1.
3Step 3: Subtract Vector \(\mathbf{v}\) from \(3\mathbf{u}\)
Perform vector subtraction by subtracting each component of \(\mathbf{v}\) from the corresponding component of \(3\mathbf{u}\):\[3\mathbf{u} - \mathbf{v} = \langle -6, -9 \rangle - \langle -3, -5 \rangle = \langle -6 - (-3), -9 - (-5) \rangle\]Simplify the expression: \[\langle -6 + 3, -9 + 5 \rangle = \langle -3, -4 \rangle\]
4Step 4: Verify the Calculation
Check each step to ensure the calculations are correct. Verify that the operations of scalar multiplication and vector subtraction were performed accurately.
Key Concepts
Scalar MultiplicationVector SubtractionPrecalculus
Scalar Multiplication
Scalar multiplication is a basic operation in vector arithmetic where we multiply a vector by a real number, known as a scalar. This operation affects each component of the vector independently, scaling it by the specified scalar. If you have a vector \( \mathbf{u} = \langle x, y \rangle \) and a scalar \( a \), the result of scalar multiplication \( a \mathbf{u} \) becomes \( \langle ax, ay \rangle \).
- Example: For the vector \( \mathbf{u} = \langle -2, -3 \rangle \), if it is multiplied by the scalar 3, each component of the vector is multiplied by 3. This results in \( 3 \times \langle -2, -3 \rangle = \langle 3 \times (-2), 3 \times (-3) \rangle = \langle -6, -9 \rangle \).
- Impact: Scalar multiplication changes the length of the vector but not its direction unless the scalar is negative, which reverses the direction.
Vector Subtraction
Vector subtraction involves subtracting one vector from another. This operation is similar to vector addition but involves reversing the direction of the vector being subtracted before performing the addition. If you have two vectors \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), their subtraction \( \mathbf{a} - \mathbf{b} \) is given by \( \langle x_1 - x_2, y_1 - y_2 \rangle \).
- Example: In the original exercise, subtracting vector \( \mathbf{v} = \langle -3, -5 \rangle \) from vector \( 3\mathbf{u} = \langle -6, -9 \rangle \) means calculating \( \langle -6 - (-3), -9 - (-5) \rangle = \langle -6 + 3, -9 + 5 \rangle = \langle -3, -4 \rangle \).
- Result: The resulting vector \( \langle -3, -4 \rangle \) signifies a new direction and magnitude based on the changes induced by the subtraction.
Precalculus
Precalculus serves as the foundation for calculus, focusing on the study and understanding of algebraic and geometric concepts. It includes topics like vectors, functions, and trigonometry, which are crucial for a smooth transition into calculus.
- Vectors in Precalculus: Vectors are an important component in precalculus, helping students understand magnitude and direction through arithmetic operations like addition, subtraction, and scalar multiplication.
- Application: Understanding vector arithmetic lays the groundwork for tackling more complex problems in calculus, such as those involving derivatives and integrals.
- Concept of Direction and Magnitude: By applying these fundamental operations to vectors, students gain insights into directional forces, which can be applied in physics and engineering.
Other exercises in this chapter
Problem 45
Use a table of values to evaluate the following limits as \(x\) increases without bound. $$\lim _{x \rightarrow \infty} \frac{5 x^{3}+2}{10 x^{3}-2 x+1}$$
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Write each statement using limit notation. As \(x\) approaches \(m\) from the left, \(f\) approaches \(L\)
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Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 16}(\sqrt{x}-16)$$
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Use a table of values to evaluate the following limits as \(x\) increases without bound. $$\lim _{x \rightarrow \infty} \frac{8 x-3 x^{2}}{7 x^{2}-x+1}$$
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