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Chapter 12: Problem 5

Find \((\) a) \(2 \mathbf{a}+\mathbf{b} .\) (b) \(3 \mathbf{a}-4 \mathbf{b},(\mathrm{c}) \mathbf{a} \cdot \mathbf{b}\). (d) \(|\mathbf{a}-\mathbf{b}|\), and (e) \(\mathbf{a} /|\mathbf{a}|\). \(\mathbf{a}=2 \mathbf{i}-\mathbf{j} . \quad \mathbf{b}=\mathbf{j}-3 \mathbf{k}\)

Short Answer

Expert verified
(a) \(4\mathbf{i} - \mathbf{j} - 3\mathbf{k}\), (b) \(6\mathbf{i} - 7\mathbf{j} + 12\mathbf{k}\), (c) -1, (d) \(\sqrt{17}\), (e) \(\frac{2}{\sqrt{5}}\mathbf{i} - \frac{1}{\sqrt{5}}\mathbf{j}\).

Step by step solution

01

Calculate 2a + b

First, calculate \(2 \mathbf{a}\) by multiplying each component of \(\mathbf{a}\) by 2. Given \(\mathbf{a} = 2\mathbf{i} - \mathbf{j}\), multiplying yields: \(2\mathbf{a} = 2(2\mathbf{i} - \mathbf{j}) = 4\mathbf{i} - 2 \mathbf{j}\). Next, add \(\mathbf{b}\): \(4\mathbf{i} - 2 \mathbf{j} + ( \mathbf{j} - 3\mathbf{k} ) = 4\mathbf{i} + (-2 + 1)\mathbf{j} - 3\mathbf{k} = 4\mathbf{i} - \mathbf{j} - 3\mathbf{k}\). The result is \(4\mathbf{i} - \mathbf{j} - 3\mathbf{k}\).
02

Calculate 3a - 4b

First, multiply \(\mathbf{a}\) by 3: \(3\mathbf{a} = 3(2\mathbf{i} - \mathbf{j}) = 6\mathbf{i} - 3\mathbf{j}\). Next, multiply \(\mathbf{b}\) by 4: \(4\mathbf{b} = 4(\mathbf{j} - 3\mathbf{k}) = 4\mathbf{j} - 12\mathbf{k}\). Then, subtract \(4\mathbf{b}\) from \(3\mathbf{a}\): \(6\mathbf{i} - 3\mathbf{j} - (4\mathbf{j} - 12\mathbf{k}) = 6\mathbf{i} - 3\mathbf{j} - 4\mathbf{j} + 12\mathbf{k} = 6\mathbf{i} - 7\mathbf{j} + 12\mathbf{k}\). The result is \(6\mathbf{i} - 7\mathbf{j} + 12\mathbf{k}\).
03

Calculate a 鈰� b

To find the dot product \(\mathbf{a} \cdot \mathbf{b}\), multiply corresponding components of \(\mathbf{a}\) and \(\mathbf{b}\) and then sum them: \(\mathbf{a} = 2\mathbf{i} - \mathbf{j}\) and \(\mathbf{b} = \mathbf{j} - 3\mathbf{k}\). Therefore, the dot product is: \((2)(0) + (-1)(1) + (0)(-3) = 0 - 1 + 0 = -1\). Thus, \(\mathbf{a} \cdot \mathbf{b} = -1\).
04

Calculate |a - b|

First, find \(\mathbf{a} - \mathbf{b}\): \((2\mathbf{i} - \mathbf{j}) - (\mathbf{j} - 3\mathbf{k}) = 2\mathbf{i} - \mathbf{j} - \mathbf{j} + 3\mathbf{k} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\). To find the magnitude, calculate the square root of the sum of the squares of the components: \(|\mathbf{a} - \mathbf{b}| = \sqrt{(2)^2 + (-2)^2 + (3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}\). Thus, \(|\mathbf{a} - \mathbf{b}| = \sqrt{17}\).
05

Calculate a / |a|

First, determine the magnitude \(|\mathbf{a}|\): \(\mathbf{a} = 2\mathbf{i} - \mathbf{j}\). The magnitude is \(|\mathbf{a}| = \sqrt{(2)^2 + (-1)^2 + (0)^2} = \sqrt{4 + 1} = \sqrt{5}\). To find \(\mathbf{a} / |\mathbf{a}|\), divide each component of \(\mathbf{a}\) by \(\sqrt{5}\): \(\mathbf{a} / |\mathbf{a}| = \frac{2\mathbf{i}}{\sqrt{5}} - \frac{1\mathbf{j}}{\sqrt{5}} = \frac{2}{\sqrt{5}}\mathbf{i} - \frac{1}{\sqrt{5}}\mathbf{j}\).

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Key Concepts

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

Dot Product
The dot product is a fundamental operation in vector algebra. It involves multiplying corresponding components of two vectors and then summing those products. For two vectors \(\mathbf{a} = ai + bj + ck\) and \(\mathbf{b} = di + ej + fk\), the dot product is calculated as:
  • \(\mathbf{a} \cdot \mathbf{b} = (a \cdot d) + (b \cdot e) + (c \cdot f)\)
The result of a dot product is a scalar, not a vector. This operation is useful in determining the angle between two vectors, as it directly relates to the cosine of the angle between them through the formula:
  • \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\)
A dot product of zero indicates that the vectors are perpendicular.
Vector Magnitude
The magnitude of a vector, often referred to as its "length" or "norm," measures its size regardless of direction. If \(\mathbf{a} = ai + bj + ck\), its magnitude \(|\mathbf{a}|\) is calculated using the formula:
  • \(|\mathbf{a}| = \sqrt{a^2 + b^2 + c^2}\)
This formula is derived from the Pythagorean theorem, extending into three-dimensional space. The magnitude is always non-negative and is zero only when the vector itself is a zero vector. Calculating the magnitude is essential for normalization, determining distances between points, and many other applications.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a real number). If you have a vector \(\mathbf{v} = ai + bj + ck\) and a scalar \(n\), the result of the scalar multiplication is:
  • \(n\mathbf{v} = (n \cdot a)i + (n \cdot b)j + (n \cdot c)k\)
This operation scales the vector by \(n\):
  • If \(|n| > 1\), the vector magnitude increases.
  • If \(0 < |n| < 1\), the vector magnitude decreases.
  • If \(n = 0\), the vector becomes a zero vector.
  • If \(n < 0\), the vector not only scales but also reverses direction.
Scalar multiplication is widely used in physics and engineering to model various phenomena like velocity and force.
Unit Vector
A unit vector is a vector with a magnitude of one. It is used to indicate direction without specifying the magnitude. To convert any vector \(\mathbf{a}\) into a unit vector \(\mathbf{\hat{a}}\), you divide the vector by its magnitude:
  • \(\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\)
This process is called normalization. If \(\mathbf{a} = ai + bj + ck\), the unit vector is:
  • \(\mathbf{\hat{a}} = \frac{a}{|\mathbf{a}|}i + \frac{b}{|\mathbf{a}|}j + \frac{c}{|\mathbf{a}|}k\)
Unit vectors are especially useful in physics to express vectors in specified directions, such as \hat{i}\, \hat{j}\, and \hat{k}\, the standard unit vectors in Cartesian coordinates.

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Most popular questions from this chapter

Describe and sketch the graphs of the equations given in Problems 1 through 30 . \(x^{2}+y^{2}-9 z^{2}=9\)

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