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Chapter 11: Problem 37

(a) Given that \(\|\mathbf{v}\|=3,\) find all values of \(k\) such that \(\|k \mathbf{v}\|=5 .\) (b) Given that \(k=-2\) and \(\|k \mathbf{v}\|=6,\) find \(\|\mathbf{v}\|\)

Short Answer

Expert verified
(a) k = 卤5/3; (b) \|\mathbf{v}\| = 3.

Step by step solution

01

Understanding Vector Magnitude Multiplication

The magnitude of a vector \(k \mathbf{v}\) is given by the product of the absolute value of the scalar \(k\) and the magnitude of the vector \(\mathbf{v}\). Therefore, \|k \mathbf{v}\| = |k| \, \|\mathbf{v}\|\. In Part (a), we know that \|k \mathbf{v}\| = 5\ and \|\mathbf{v}\| = 3\. We will use this relationship to find \(k\).
02

Solving for k in Part (a)

Use the equation \|k\mathbf{v}\| = |k|\|\mathbf{v}\|\. Substitute the known values: \(5 = |k| \times 3\). Solve for \( |k| \): \[ |k| = \frac{5}{3} \] This means that \( k \) can be either \( \frac{5}{3} \) or \( -\frac{5}{3} \).
03

Part (a) Conclusion

The possible values of \(k\) such that \|k \mathbf{v}\| = 5\ are \( \frac{5}{3} \) and \( -\frac{5}{3} \).
04

Understanding Given Values in Part (b)

We know \( k = -2 \) and \|k \mathbf{v}\| = 6\. We need to find \|\mathbf{v}\|\. Use the formula \|k \mathbf{v}\| = |k| \|\mathbf{v}\|\ to determine \(\|\mathbf{v}\|\).
05

Solving for Magnitude of v in Part (b)

Given \( |k| = 2 \) and \|k \mathbf{v}\| = 6\, the equation becomes \(6 = 2\|\mathbf{v}\|\). Solve for \(\|\mathbf{v}\|\): \[\|\mathbf{v}\| = \frac{6}{2} = 3\]
06

Part (b) Conclusion

The magnitude of the vector \(\mathbf{v}\|\), when \(k = -2\), is 3.

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Key Concepts

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

Scalar Multiplication
Scalar multiplication in vectors refers to the process of multiplying a vector by a numerical value, called a scalar. This operation affects only the magnitude of the vector, not its direction. When you multiply a vector \( \mathbf{v} \) by a scalar \( k \), you scale the vector in proportion to \( k \).
  • If \( k \) is positive, the direction of the vector remains unchanged, but the magnitude is scaled.
  • If \( k \) is negative, the vector is not only scaled but also reflected (flipped) in the opposite direction.
  • If \( k \) equals zero, the vector becomes the zero vector, losing its direction entirely.
For example, if \( \|\mathbf{v}\| = 3 \) and you multiply it by a scalar \( k \) to achieve \( \|k \mathbf{v}\| = 5 \), you solve for \( k \) to determine the factors that will scale \( \mathbf{v} \) appropriately (i.e., \( k = \frac{5}{3} \) or \( k = -\frac{5}{3} \)). This illustrates how scalar multiplication expands or shrinks the magnitude of vectors.
Vector Magnitudes
The vector magnitude, or length, is a measure of how long a vector is. It can be visualized as the distance from the vector's tail to its head. Calculating the magnitude of a vector \( \mathbf{v} \) is straightforward using the formula based on its components:
  • If \( \mathbf{v} = (x, y, z) \), then the magnitude is \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \).
  • This results in a non-negative scalar value that represents the size of the vector.
In exercises, such as finding all \( k \) for which \( \|k \mathbf{v}\| = 5 \) when \( \|\mathbf{v}\| = 3 \), these principles are applied. By using the relationship \( \|k \mathbf{v}\| = |k| \|\mathbf{v}\| \), you solve for \( k \) to ensure the vector reaches the desired magnitude. Calculating vector magnitudes is crucial for understanding the physical impact of vectors in real-world applications.
Vector Solutions
Solving vector problems often requires a step-by-step approach to ensure accuracy.
  • Identify the given information and what needs to be found.
  • Apply relevant vector equations to the problem.
  • Rearrange equations to isolate the unknown variable.
In Part (a) of this exercise, you are given \( \|\mathbf{v}\| \) and \( \|k \mathbf{v}\| \), and asked to find \( k \). Substitute the known magnitudes into the equation \( \|k \mathbf{v}\| = |k| \|\mathbf{v}\| \) and solve for \( |k| \). Recognize that \( k \) can be either positive or negative, given its absolute value. Part (b) follows a similar process of inserting known values into the formula, this time solving for \( \|\mathbf{v}\| \). Efficiently solving vector problems involves systematically applying these steps to achieve clear and precise solutions.

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Most popular questions from this chapter

Convert from rectangular to cylindrical coordinates. $$ \begin{array}{ll}{\text { (a) }(\sqrt{2},-\sqrt{2}, 1)} & {\text { (b) }(0,1,1)} \\ {\text { (c) }(-4,4,-7)} & {\text { (d) }(2,-2,-2)}\end{array} $$

Convert from spherical to rectangular coordinates. $$ \begin{array}{ll}{\text { (a) }(5, \pi / 6, \pi / 4)} & {\text { (b) }(7,0, \pi / 2)} \\ {\text { (c) }(1, \pi, 0)} & {\text { (d) }(2,3 \pi / 2, \pi / 2)}\end{array} $$

Determine whether the statement is true or false. Explain your answer. The graph of \(r=f(\theta)\) in cylindrical coordinates can always be obtained by extrusion of the polar graph of \(r=f(\theta)\) in the \(x y\) -plane.

Show that the lines $$ \begin{array}{ll}{x=-2+t,} & {y=3+2 t, \quad z=4-t} \\ {x=3-t,} & {y=4-2 t, \quad z=t}\end{array} $$ are parallel and find an equation of the plane they determine.

An equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. $$ x^{2}+y^{2}+z^{2}=9 $$
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