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Chapter 9: Problem 4

If \(\mathbf{a}=\mathbf{i}+8 \mathbf{j}+\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+10 \mathbf{j}+\mathbf{k},\) find a unit vector with positive first coordinate orthogonal to both a and \(\mathrm{b}\). ________\(\mathbf{i}+\)_________\(\mathbf{j}+$$\mathbf{k}\)

Short Answer

Expert verified
\(\hat{\mathbf{c}} = \frac{-2}{\sqrt{53}} \mathbf{i} + \frac{7}{\sqrt{53}} \mathbf{k}\)

Step by step solution

01

Write down the given vectors

Let vectors \(\mathbf{a} = \mathbf{i} + 8 \mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + 10 \mathbf{j} + \mathbf{k}\).
02

Calculate the cross product of a and b

The cross product of two vectors can be calculated using the determinant formula: \(\mathbf{c} = \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 8 & 1 \\ 1 & 10 & 1 \end{vmatrix}\) Expanding the determinant: \(\mathbf{c} = \mathbf{i}(8-10) - \mathbf{j}(1-1) + \mathbf{k}(1-8)\) \(\mathbf{c} = -2 \mathbf{i} + 7 \mathbf{k}\)
03

Find the magnitude of c

Now, we need to find the magnitude of vector \(\mathbf{c}\): \(|\mathbf{c}| = \sqrt{(-2)^2 + (7)^2} = \sqrt{53}\)
04

Find the unit vector of c

To find the unit vector of \(\mathbf{c}\), we'll divide each component by the magnitude: \(\hat{\mathbf{c}} = \frac{\mathbf{c}}{|\mathbf{c}|} = \frac{-2 \mathbf{i} + 7 \mathbf{k}}{\sqrt{53}}\) Since the first coordinate is positive, this is the unit vector orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}:\) \(\hat{\mathbf{c}} = \frac{-2}{\sqrt{53}} \mathbf{i} + \frac{7}{\sqrt{53}} \mathbf{k}\)

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Key Concepts

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

Cross Product
The cross product is a unique operation available for three-dimensional vectors. It helps in finding a vector that is orthogonal, or perpendicular, to two given vectors. When you have two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), their cross product, \( \mathbf{a} \times \mathbf{b} \), results in another vector that is at right angles to both \( \mathbf{a} \) and \( \mathbf{b} \).

Here's how the cross product is calculated:
  • Firstly, arrange the vectors in a determinant form with the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as the first row.
  • Second and third rows contain the components of the vectors \( \mathbf{a} \) and \( \mathbf{b} \).
  • Expand the determinant to derive the vector orthogonal to both.
The great thing about the cross product is that it efficiently solves problems involving the direction of forces, rotational speeds, and more in physics and engineering.
Unit Vector
A unit vector is a vector with a magnitude of 1. It keeps the direction of the original vector but scales its length down to one unit. This is particularly useful in various calculations, including normalizing vectors to ensure consistent measurements.

To convert any vector into a unit vector, you divide each of its components by its magnitude. The formula for finding a unit vector \( \hat{\mathbf{c}} \) from vector \( \mathbf{c} \) is:
  • Calculate the magnitude of \( \mathbf{c} \).
  • Divide each component of the vector by this magnitude.
The simple mathematical expression for this is \( \hat{\mathbf{c}} = \frac{\mathbf{c}}{|\mathbf{c}|} \).

Unit vectors are commonly used to represent directions in a coordinate system, making them an essential foundational concept in vector calculus.
Orthogonal Vectors
In vector calculus, vectors are orthogonal if they meet at a right angle, meaning the dot product of the two vectors is zero. When two vectors are orthogonal, they are essentially perpendicular, not pointing in the same or opposite directions.

Finding vectors that are orthogonal is significant in numerous applications, such as verifying solutions to vector problems or in optimizations where minimal overlap is desired between directions, like in projections.

Consider \( \mathbf{a} \) and \( \mathbf{b} \):
  • The condition for orthogonality can be checked by ensuring \( \mathbf{a} \cdot \mathbf{b} = 0 \). If they are, indeed, zero, then these vectors are at 90 degrees to each other.
In practical applications such as computer graphics or physics, orthogonal vectors are in symphony, providing balanced projections or decompositions of forces.
Magnitude of a Vector
The magnitude of a vector, often described as the length or size of the vector, measures how much space the vector spans. It is especially fundamental in deciding the norm or scale of a vector in computations.

Here's how to determine the magnitude for a vector, say \( \mathbf{c} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \):
  • Square each of the vector's components.
  • Add all those squared values up.
  • Calculate the square root of the sum.
Mathematically, this is represented as \( |\mathbf{c}| = \sqrt{a^2 + b^2 + c^2} \).

The magnitude provides valuable insights into understanding the physical representation of vectors. It answers how far the vector extends in mathematical and physical spaces, which is crucial for applications in motion, speed, and other vector dynamics.

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Most popular questions from this chapter

Consider the function \(h\) defined by \(h(x, y)=8-\sqrt{4-x^{2}-y^{2}}\). a. What is the domain of \(h ?\) (Hint: describe a set of ordered pairs in the plane by explaining their relationship relative to a key circle.) b. The range of a function is the set of all outputs the function generates. Given that the range of the square root function \(g(t)=\sqrt{t}\) is the set of all nonnegative real numbers, what do you think is the range of \(h ?\) Why? c. Choose 4 different values from the range of \(h\) and plot the corresponding level curves in the plane. What is the shape of a typical level curve? d. Choose 5 different values of \(x\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). e. Choose 5 different values of \(y\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). f. Sketch an overall picture of the surface generated by \(h\) and write at least one sentence to describe how the surface appears visually. Does the surface remind you of a familiar physical structure in nature?

Let \(\mathbf{u}=\langle 2,1\rangle\) and \(\mathbf{v}=\langle 1,2\rangle\). a. Determine the components and draw geometric representations of the vectors \(2 \mathbf{u}, \frac{1}{2} \mathbf{u},(-1) \mathbf{u},\) and (-3) \(\mathbf{u}\) on the same set of axes. b. Determine the components and draw geometric representations of the vectors \(\mathbf{u}+\mathbf{v}, \mathbf{u}+2 \mathbf{v},\) and \(\mathbf{u}+3 \mathbf{v}\) on the same set of axes. c. Determine the components and draw geometric representations of the vectors \(\mathbf{u}-\mathbf{v}, \mathbf{u}-2 \mathbf{v},\) and \(\mathbf{u}-3 \mathbf{v}\) on the same set of axes. d. Recall that \(\mathbf{u}-\mathbf{v}=\mathbf{u}+(-1) \mathbf{v}\). Sketch the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(\mathbf{u}-\mathbf{v}\) on the same set of axes. Use the "tip to tail" perspective for vector addition to explain the geometric relationship between \(\mathbf{u}, \mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\)

Let \(\mathbf{a}=\langle 2,3,3\rangle\) and \(\mathbf{b}=\langle 0,1,4\rangle\). Compute: \(\mathbf{a}+\mathbf{b}=(\)________,________,________) \(\mathbf{a}-\mathbf{b}=(\)________,________,________) \(2 \mathbf{a}=(\underline{\mathbf{a}}+4 \mathbf{b}=(\)________,________,________) \(3 \mathbf{a}+=\)___________

Consider the standard helix parameterized by \(\mathbf{r}(t)=\cos (t) \mathbf{i}+\sin (t) \mathbf{j}+t \mathbf{k}\). a. Recall that the unit tangent vector, \(\mathbf{T}(t),\) is the vector tangent to the curve at time \(t\) that points in the direction of motion and has length 1. Find \(\mathbf{T}(t)\). b. Explain why the fact that \(|\mathbf{T}(t)|=1\) implies that \(\mathbf{T}\) and \(\mathbf{T}^{\prime}\) are orthogonal vectors for every value of \(t .\) (Hint: note that \(\mathbf{T} \cdot \mathbf{T}=\) \(|\mathbf{T}|^{2}=1,\) and compute \(\left.\frac{d}{d t}[\mathbf{T} \cdot \mathbf{T}] .\right)\) c. For the given function \(\mathbf{r}\) with unit tangent vector \(\mathbf{T}(t)\) (from (a)), determine \(\mathbf{N}(t)=\frac{1}{\left|\mathbf{T}^{\prime}(t)\right|} \mathbf{T}^{\prime}(t)\) d. What geometric properties does \(\mathbf{N}(t)\) have? That is, how long is this vector, and how is it situated in comparison to \(\mathbf{T}(t) ?\) e. Let \(\mathbf{B}(t)=\mathbf{T}(t) \times \mathbf{N}(t),\) and compute \(\mathbf{B}(t)\) in terms of your results in (a) and (c). f. What geometric properties does \(\mathbf{B}(t)\) have? That is, how long is this vector, and how is it situated in comparison to \(\mathbf{T}(t)\) and \(\mathbf{N}(t) ?\) g. Sketch a plot of the given helix, and compute and sketch \(\mathbf{T}(\pi / 2)\), \(\mathbf{N}(\pi / 2),\) and \(\mathbf{B}(\pi / 2)\)

In this exercise we verify the curvature formula $$ \kappa=\frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|^{3}} $$ a. Explain why $$ \left|\mathbf{r}^{\prime}(t)\right|=\frac{d s}{d t} $$ b. Use the fact that \(\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}\) and \(\left|\mathbf{r}^{\prime}(t)\right|=\frac{d s}{d t}\) to explain why $$ \mathbf{r}^{\prime}(t)=\frac{d s}{d t} \mathbf{T}(t) $$ c. The Product Rule shows that $$ \mathbf{r}^{\prime \prime}(t)=\frac{d^{2} s}{d t^{2}} \mathbf{T}(t)+\frac{d s}{d t} \mathbf{T}^{\prime}(t) $$ Explain why $$ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)=\left(\frac{d s}{d t}\right)^{2}\left(\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right) $$ d. In Exercise 9.8 .5 .14 we showed that \(|\mathbf{T}(t)|=1\) implies that \(\mathbf{T}(t)\) is orthogonal to \(\mathbf{T}^{\prime}(t)\) for every value of \(t\). Explain what this tells us about \(\left|\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right|\) and conclude that $$ \left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|=\left(\frac{d s}{d t}\right)^{2}\left|\mathbf{T}^{\prime}(t)\right| $$ e. Finally, use the fact that \(\kappa=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|}\) to verify that $$ \kappa=\frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|^{3}} $$
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