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Chapter 11: Problem 39

A vector \(\mathbf{u}=2 \mathbf{i}+3 \mathbf{j}+z \mathbf{k}\) emanating from the origin points into the first octant (i.e., that part of three-space where all components are positive). If \(\|\mathbf{u}\|=5\), find \(z\).

Short Answer

Expert verified
The value of \(z\) is \(2\sqrt{3}\).

Step by step solution

01

Understanding the First Octant

A vector points into the first octant if all of its components are positive. This means that the components for \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) must all be greater than zero. Since \(\mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + z \mathbf{k}\), it follows that \(z > 0\).
02

Applying the Magnitude Formula

The magnitude (or length) of a vector \(\mathbf{u} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\) in 3-dimensional space is given by the formula:\[\|\mathbf{u}\| = \sqrt{a^2 + b^2 + c^2}\]For \(\mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + z \mathbf{k}\), the magnitude is 5, thus:\[\sqrt{2^2 + 3^2 + z^2} = 5\]
03

Solving for \(z\)

First, simplify the equation:\[\sqrt{4 + 9 + z^2} = 5\]This simplifies to:\[\sqrt{13 + z^2} = 5\]Square both sides to eliminate the square root:\[13 + z^2 = 25\]Subtract 13 from both sides to isolate \(z^2\):\[z^2 = 12\]Take the square root of both sides, considering only the positive solution because \(z > 0\):\[z = \sqrt{12} = 2\sqrt{3}\]
04

Ensuring Solution is Valid

Reassess the conditions: \(z = 2\sqrt{3}\) is positive, satisfying the requirement that \(z > 0\). The computation is verified since its square would return to \(z^2 = 12\), confirming consistency with the magnitude condition.

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Key Concepts

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

Magnitude of a Vector
In vector calculus, understanding the magnitude of a vector is key when analyzing and solving vector problems. The magnitude, often referred to as the length, measures how long a vector is in space. It's computed using the Pythagorean theorem extended to three dimensions. For a vector
  • ewline \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is calculated with \( \|\mathbf{u}\| = \sqrt{a^2 + b^2 + c^2} \).
In our example, the vector \( \mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + z \mathbf{k} \) has a known magnitude of 5.ewline By applying the magnitude formula, substituting the given components, we form an equation: \[ \sqrt{2^2 + 3^2 + z^2} = 5 \].ewline This setup allows us to solve for \( z \), using algebraic steps to isolate and compute \( z \) accurately. Computing magnitude correctly is crucial because it tells us how to place the vector in space accurately.
Three-Dimensional Space
Vectors exist in a space that has three dimensions, often represented by the x, y, and z axes, corresponding to \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) unit vectors respectively. Understanding this space helps us visualize where vectors "live" and how they interact. Each dimension adds a degree of freedom to movement:
  • ewline X-axis: Left or right
  • Y-axis: Up or down
  • Z-axis: Forward or backward
When working with vectors in this space, especially for computation or physics applications, we see how they could point in any direction, represented by a set of coordinates. In our exercise, the vector starts from the origin point (\(0,0,0\)).ewline This means all calculations begin at a defined central point, moving outward based on the vector's components. Grasping these spatial relationships allows us to fully capture the vector's behavior and effect, especially when considering the effects of forces, directions, or transformations within a three-dimensional context.
First Octant
The first octant is a specific region within three-dimensional Cartesian space. It's defined where all three coordinates are positive. Picture the 3D coordinate system divided into eight parts by the three axes.ewline The first octant occupies the positive space of each:
  • Positive x (\( \mathbf{i} \))
  • Positive y (\( \mathbf{j} \))
  • Positive z (\( \mathbf{k} \))
Our exercise highlights this: the vector \( \mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + z \mathbf{k} \) must have positive components. Hence, for \( z \), we know \( z > 0 \). Ensuring each component's positivity confirms the vector rightly points into this specific quadrant.ewline Understanding this is important in real-world scenarios where direction and orientation matter, such as navigation, engineering, and physics problems. By acknowledging the first octant's boundaries, we know precisely where in space our solutions and vectors will operate.

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Most popular questions from this chapter

. Sketch the path for a particle if its position vector is \(\mathbf{r}=\sin t \mathbf{i}+\sin 2 t \mathbf{j}, 0 \leq t \leq 2 \pi\) (you should get a figure eight). Where is the acceleration zero? Where does the acceleration vector point to the origin?

Evaluate the integrals $$ \int_{0}^{1}\left(e^{\prime} \mathbf{i}+e^{-t \mathbf{j}}\right) d t $$

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

Find the equation of the surface that results when the curve \(4 x^{2}+3 y^{2}=12\) in the \(x y\) -plane is revolved about the \(y\) -axis.

If the curve \(z=x^{2}\) in the \(x z\) -plane is revolved about the \(z\) -axis, the resulting surface has equation \(z=x^{2}+y^{2}\), obtained as a result of replacing \(x\) by \(\sqrt{x^{2}+y^{2}}\). If \(y=2 x^{2}\) in the \(x y\) -plane is revolved about the \(y\) -axis, what is the equation of the resulting surface?
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