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Chapter 12: Problem 36

Give a geometric description of the following sets of points. $$x^{2}+y^{2}+z^{2}-8 x+14 y-18 z \geq 65$$

Short Answer

Expert verified
Based on the solution above, the geometric description of the set of points that satisfy the inequality $$x^{2}+y^{2}+z^{2}-8 x+14 y-18 z \geq 65$$ represents a sphere and its exterior. The sphere has a center at point C(4, -7, 9) and a radius of \(\sqrt{211}\).

Step by step solution

01

Complete the square for each variable

Let's start by completing the square for the x, y, and z terms separately. For x: $$x^2 - 8x = (x - 4)^2 - 16$$ For y: $$y^2 + 14y = (y + 7)^2 - 49$$ For z: $$z^2 - 18z = (z - 9)^2 - 81$$ Now, substitute these back into the inequality: $$(x - 4)^2 - 16 + (y + 7)^2 - 49 + (z - 9)^2 - 81 \geq 65$$
02

Simplify the inequality

Combine the constants on the left side of the inequality: $$(- 16) + (- 49) + (- 81) + (x - 4)^2 + (y + 7)^2 + (z - 9)^2 \geq 65$$ Add 146 to both sides of the inequality: $$(x-4)^2 + (y+7)^2 + (z-9)^2 \geq 211$$
03

Identify the geometric shape and properties

We can now see that the inequality represents the equation for a sphere in 3D space: $$(x - a)^2 + (y - b)^2 + (z - c)^2 \geq r^2$$ The inequality being greater than or equal to \(r^2\) means that the set of points will be the points on the surface of the sphere and all the points outside of the sphere. In this case, \(a = 4\), \(b = -7\), \(c = 9\), and \(r = \sqrt{211}\). So, the center of the sphere is at point C(4, -7, 9) and the radius is \(\sqrt{211}\).
04

Geometric Description

The given set of points represents a sphere centered at point C(4, -7, 9) with a radius of \(\sqrt{211}\), including the points on the surface and all the points outside the sphere.

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Most popular questions from this chapter

Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-1,2,3\rangle, \mathbf{v}=\langle 2,1,1\rangle\)

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot \mathbf{v}=\mathbf{v} \cdot \mathbf{u}\)

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.

Find the points (if they exist) at which the following planes and curves intersect. $$y=1 ; \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$
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