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Linear combinations A sum of scalar multiples of two or more vectors (such as \(c_{1} \mathbf{u}+c_{2} \mathbf{v}+c_{3} \mathbf{w},\) where \(c_{i}\) are scalars) is called a linear combination of the vectors. Let \(\mathbf{i}=\langle 1,0\rangle, \mathbf{j}=\langle 0,1\rangle\) \(\mathbf{u}=\langle 1,1\rangle,\) and \(\mathbf{v}=\langle-1,1\rangle\) Express \langle 4,-8\rangle as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\)

Short Answer

Expert verified
Question: Express the vector 鉄4, -8鉄 as a linear combination of the given vectors 饾惍 = 鉄1, 1鉄 and 饾惎 = 鉄-1, 1鉄. Answer: 鉄4, -8鉄 = (-2)饾惍 + (-6)饾惎

Step by step solution

01

Set up the linear equations

Based on the linear combination definition, we have the following equation: 鉄4, -8鉄 = c1鉄1, 1鉄 + c2鉄-1, 1鉄 Comparing the coordinates of the vectors, we get the following system of linear equations: `1*c1 + (-1)*c2 = 4` `1*c1 + 1*c2 = -8`
02

Solve the system of linear equations

Now, we need to solve this system to find the values of c1 and c2. We can either use substitution or elimination method. Here we use elimination: Add the two equations: `1*c1 + (-1)*c2 + 1*c1 + 1*c2 = 4 - 8` `2*c1 = -4` Now, divide by 2: c1 = -2 Substitute the value of c1 in one of the initial equations, e.g., the first one: `1*(-2) + (-1)*c2 = 4` `-2 - c2 = 4` Add 2 to both sides: `-c2 = 6` Multiply by -1: c2 = -6
03

Write the linear combination

Now that we found the values of c1 and c2, we can express the vector 鉄4, -8鉄 as a linear combination of 饾惍 and 饾惎: 鉄4, -8鉄 = (-2)饾惍 + (-6)饾惎 So, the vector 鉄4, -8鉄 can be expressed as a linear combination of the given vectors 饾惍 and 饾惎 with scalar coefficients -2 and -6 respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vectors
Vectors are essential in mathematics and physics and can be visualized as arrows with both magnitude and direction. A vector is typically represented in a coordinate space using angle brackets, like \(<1, 1>\). These components show the vector's influence in each dimension of space. In a 2D space, vectors are expressed using vertical and horizontal components. Vectors can be added together, and they can scale based on scalar multiplication. In our example, \(\mathbf{u} = \langle 1,1 \rangle\) and \(\mathbf{v} = \langle -1,1 \rangle\) are two vectors. By applying different scalar values to these vectors, we can form a linear combination that results in another vector. This process allows for the creation of many other vectors by combining existing ones.
  • A vector's representation depends on the space dimension it belongs to. In 3D, for instance, a vector might be \(<1, 0, -1>\).
  • Vectors are used extensively to solve real-world problems, like determining forces, velocities, and more.
Linear Equations
Linear equations form the backbone of algebra and are equations of the first degree. They usually appear in the form \((Ax + By = C)\), where \(A\), \(B\), and \(C\) are constants. Solving linear equations involves finding the values of the unknowns that satisfy the equation. In the context of the exercise, involving linear combinations of vectors, you are tasked with solving a system of linear equations to find suitable scalars that multiply the given vectors to arrive at the target vector. Our system was:
  • \(1 \cdot c_1 - 1 \cdot c_2 = 4\)
  • \(1 \cdot c_1 + 1 \cdot c_2 = -8\)
The elimination method, a popular technique to solve these equations, involves combining equations to cancel one of the variables, allowing for simpler calculations.
  • The method used typically depends on the number of variables and equations. Substitution is another method often used.
  • Graphically, linear equations can represent lines in a coordinate plane, and their intersection is the solution to the system.
Scalars
Scalars are simple but powerful mathematical values that help us in numerous calculations, especially in vector mathematics. Unlike vectors, scalars consist of magnitude only and have no direction. Common examples of scalars include numbers representing quantities such as temperature, speed, or mass. In the context of vectors, scalars are crucial because they allow vector scaling. By multiplying a vector by a scalar, you modify the vector's magnitude without altering its direction. This is a foundational concept in forming linear combinations of vectors. Given the scalars found, \(c_1 = -2\) and \(c_2 = -6\), we multiply them by \(\mathbf{u}\) and \(\mathbf{v}\) to adjust their influence and achieve the desired resultant vector.
  • Scalars can be positive or negative, influencing whether vectors retain or reverse their direction.
  • The process of obtaining new vectors by scalar multiplication is fundamental in fields like physics, where changing magnitudes are crucial.

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Most popular questions from this chapter

Use vectors to show that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) is the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) .\) (Hint: Let \(O\) be the origin and let \(M\) be the midpoint of \(P Q .\) Draw a picture and show that \( {O M}= {O P}+\frac{1}{2} {P Q}= {O P}+\frac{1}{2}( {O Q}- {O P})\)

Four-cable load A 500 -Ib load hangs from four cables of equal Iength that are anchored at the points (卤2,0,0) and (0,卤2,0) The load is located at \((0,0,-4) .\) Find the vectors describing the forces on the cables due to the load. (graph cannot copy)

Assume \( {P Q}\) equals \( {R S} .\) Does it follow that \( {P R}\) is equal to \( {Q S}\) ? Prove your conclusion.

Linear combinations A sum of scalar multiples of two or more vectors (such as \(c_{1} \mathbf{u}+c_{2} \mathbf{v}+c_{3} \mathbf{w},\) where \(c_{i}\) are scalars) is called a linear combination of the vectors. Let \(\mathbf{i}=\langle 1,0\rangle, \mathbf{j}=\langle 0,1\rangle\) \(\mathbf{u}=\langle 1,1\rangle,\) and \(\mathbf{v}=\langle-1,1\rangle\) For arbitrary real numbers \(a\) and \(b\), express \(\langle a, b\rangle\) as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\)

Determine whether the following statements are true using a proof or counterexample. Assume \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=\mathbf{w} \cdot(\mathbf{u} \times \mathbf{v})$$

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