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Find \((a) \mathbf{u} \cdot \mathbf{v},(b) \mathbf{u} \cdot \mathbf{u},(c)\|\mathbf{u}\|^{2},(d)(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and (e) \(\mathbf{u} \cdot(5 \mathbf{v})\). \(\mathbf{u}=(4,0,-3,5), \quad \mathbf{v}=(0,2,5,4)\)

Short Answer

Expert verified
The solutions are (a) -7, (b) 41, (c) 41, (d) (0,-14,-35,-28), (e) -35.

Step by step solution

01

Calculate the dot product of vectors u and v

To calculate the dot product of two vectors, you multiply corresponding elements of both vectors together and then sum them up. The dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as \((4*0)+(0*2)+(-3*5)+(5*4) = -7\).
02

Calculate the dot product of vector u with itself

The dot product of a vector with itself is just the sum of the squares of its elements. The dot product \(\mathbf{u} \cdot \mathbf{u}\) is calculated as \((4*4)+(0*0)+(-3*-3)+(5*5) = 41\).
03

Calculate the square of the magnitude of vector u

The square of the magnitude of a vector is the same as the dot product of the vector with itself. Therefore, \(\|\mathbf{u}\|^{2}\) is also 41.
04

Calculate the dot product times vector v

For this we multiply each element of vector v by the dot product calculated in step 1. Therefore, \((\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) is \(-7*(0,2,5,4) = (0,-14,-35,-28)\).
05

Calculate the dot product of vector u and 5 times vector v

The dot product is distributive over vector addition, we can pull constants out of the dot product. Therefore, we can solve this by multiplying the dot product from step 1 by 5. So, \(\mathbf{u} \cdot(5 \mathbf{v})\)= 5*(-7) = -35.

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Key Concepts

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

Vector Operations
Vector operations are fundamental in the study of linear algebra and have various applications in physics, engineering, and computer science. Understanding vector operations is crucial in solving problems related to direction and magnitude in multi-dimensional spaces. One common operation is the dot product, also known as the scalar product, which takes two vectors and returns a single number (scalar). This operation is performed by multiplying corresponding components of the two vectors and then summing the products. Mathematically, for two vectors \(\mathbf{a}=(a_1, a_2, ... , a_n)\) and \(\mathbf{b}=(b_1, b_2, ... , b_n)\), the dot product is given by \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n\).

In our textbook exercise, the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\) was calculated by following this method. Other vector operations include vector addition, subtraction, and cross product. Each has its own geometric and algebraic interpretation, which students must grasp to solve more complex problems involving vectors.
Magnitude of a Vector
The magnitude of a vector, often denoted as \(\|\mathbf{v}\|\), is a measure of its length and is always a non-negative scalar value. For a vector in a coordinate space, the magnitude can be visualized as the straight-line distance from the origin to the point defined by the vector. To calculate it, we use the Pythagorean theorem in multi-dimensional space. Specifically, for a vector \(\mathbf{v}=(v_1, v_2, ... , v_n)\), its magnitude is \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2}\).

In our exercise, the magnitude squared of vector \(\mathbf{u}\), or \(\|\mathbf{u}\|^2\), is found by squaring each component of \(\mathbf{u}\) and adding them, which simplifies the square root operation. This process is directly related to the dot product of \(\mathbf{u}\) with itself, yielding the result of 41.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a single number), which scales the vector's magnitude without affecting its direction if the scalar is positive. For a given scalar \(c\) and vector \(\mathbf{v} = (v_1, v_2, ... , v_n)\), scalar multiplication is defined by \(c\mathbf{v} = (cv_1, cv_2, ... , cv_n)\).

This operation is showcased in parts (d) and (e) of the exercise. In part (d), we multiplied each component of \(\mathbf{v}\) by the previously calculated dot product, which demonstrates how the vector scales when multiplied by a scalar. In part (e), the scalar multiplication principle allows us to simplify the problem by first multiplying the scalar with the dot product, because of the distributive property of scalar and vector multiplication. As a result, the dot product of \(\mathbf{u}\) with \(5\mathbf{v}\) is the same as 5 times the dot product of \(\mathbf{u}\) with \(\mathbf{v}\), leading to a solution of -35.

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Most popular questions from this chapter

Find \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) .\) This quantity is called the triple scalar product of \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\). $$\mathbf{u}=\mathbf{i}, \quad \mathbf{v}=\mathbf{j}, \quad \mathbf{w}=\mathbf{k}$$

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The table shows the sales \(y\) (in millions of dollars) for Dell Incorporated during the years 1996 to 2007 . Find the least squares regression line and the least squares cubic regression polynomial for the data. Let \(t\) represent the year, with \(t=-4\) corresponding to \(1996 .\) Which model is the better fit for the data? Why? (Source: Dell Inc.) $$\begin{aligned} &\begin{array}{l|llll} \hline \text {Year} & 1996 & 1997 & 1998 & 1999 \\ \text {Sales, } y & 7759 & 12,327 & 18,243 & 25,265 \\ \hline \text {Year} & 2000 & 2001 & 2002 & 2003 \\ \text {Sales, } y & 31,888 & 31,168 & 35,404 & 41,444 \\ \hline \end{array}\\\ &\begin{array}{l|llll} \hline \\ \hline \text {Year} & 2004 & 2005 & 2006 & 2007 \\ \text {Sales, } y & 49,205 & 55,908 & 58,200 & 61,000 \\ \hline \end{array} \end{aligned}$$

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