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Chapter 12: Problem 30

Give a geometric description of the following sets of points. $$(x+1)^{2}+y^{2}+z^{2}-2 y-24=0$$

Short Answer

Expert verified
Answer: The set of points represent a sphere with center (-1, 1, 0) and radius 5.

Step by step solution

01

Identify the shape

The given equation is $$(x+1)^{2}+y^{2}+z^{2}-2 y-24=0.$$ First, we need to rewrite it in a more recognizable form. We do this by completing the square for the \(y\) term.
02

Complete the square

To complete the square for the \(y\) term, we need to add and subtract a value such that we form a perfect square trinomial for the \(y\) term. We see that \(y^2 - 2y\) can form a perfect square trinomial if we add \(1\). Therefore, we add and subtract \(1\) in the equation: $$(x+1)^{2}+y^{2}+z^{2}-2 y-24+1-1=0.$$ This can be rearranged as $$(x+1)^{2}+y^{2}+z^{2}-2 y+1-25=0.$$So, we get $$(x+1)^{2}+(y-1)^{2}+z^{2}=25.$$
03

Identify the geometric shape

Now the equation has the form $$(x+1)^{2}+(y-1)^{2}+z^{2}=25.$$ We can see this equation represents a sphere in 3D space because it matches the standard equation of a sphere: $$(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2,$$ where \((a, b, c)\) is the center and \(r\) is the radius of the sphere.
04

Identify the center and the radius

Comparing the given equation with the standard equation of a sphere, we find: $$a=-1, b=1, c=0,$$ and $$r^2=25.$$ So, the center of the sphere is \((-1, 1, 0)\), and since \(r^2=25\), the radius is \(r=5\).
05

Final Geometric Description

The geometric description of the given set of points is a sphere with center \((-1, 1, 0)\) and radius \(5\).

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Most popular questions from this chapter

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{array}{l} \mathbf{r}(t)=\langle 4+t,-2 t, 1+3 t\rangle ;\\\ \mathbf{R}(s)=\langle 1-7 s, 6+14 s, 4-21 s\rangle \end{array}$$

Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left(x_{0}, y_{0}, z_{0}\right\rangle, \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why this function describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Evaluate the following limits. $$\lim _{t \rightarrow 2}\left(\frac{t}{t^{2}+1} \mathbf{i}-4 e^{-t} \sin \pi t \mathbf{j}+\frac{1}{\sqrt{4 t+1}} \mathbf{k}\right)$$

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors in \(\mathbb{R}^{3}\). a. Prove that the equation \(\mathbf{u} \times \mathbf{z}=\mathbf{v}\) has a nonzero solution \(\mathbf{z}\) if and only if \(\mathbf{u} \cdot \mathbf{v}=0 .\) (Hint: Take the dot product of both sides with v.) b. Explain this result geometrically.
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