Answered: Given a ray r(t) = (0; 0; 0) + t(1; 0;…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter10: Analytic Geometry

Section10.6: The Three-dimensional Coordinate System

Problem 41E: Does the sphere x2+y2+z2=100 have symmetry with respect to the a x-axis? b xy-plane?

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Question

Given a ray r(t) = (0; 0; 0) + t(1; 0; 0), t ≥ 0, and a set of spheres of unitradius and centered respectively at: (1) O = (0; 0; 0), (2) O = (3; 0; 0), (3) O = (1; 1; 0), (4) O = (-3; 0; 0), (5) O = (0; 3; 0). Which of the given spheres will be intersected from outside by the ray?

Expert Solution

Step 1

Given ray is $r (t) = (0, 0, 0) + t (1, 0, 0)$ , where $t \geq 0$ .

The given ray can be written as:

$\begin{array}{rcl} r (t) & = & (0, 0, 0) + t (1, 0, 0) \\ (x, y, z) & = & (t, 0, 0) \end{array}$

Therefore any point on the ray $r (t) = (0, 0, 0) + t (1, 0, 0)$ will of the form of $(t, 0, 0)$ for $t \geq 0$ .

That is all the non negative points of $x -$ axis.

(1).

Consider the sphere of unit radius and centered at $(0, 0, 0)$ .

Therefore its equation can be written as:

$\begin{array}{rcl} {(x - 0)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2} & = & 1^{2} \\ x^{2} + y^{2} + z^{2} & = & 1 \end{array}$

Observe that the ray $r (t) = (0, 0, 0) + t (1, 0, 0)$ started from origin and along $x -$ axis, and the sphere $\begin{array}{rcl} x^{2} + y^{2} + z^{2} & = & 1 \end{array}$ is also centered at origin with unit radius. Hence the sphere $\begin{array}{rcl} x^{2} + y^{2} + z^{2} & = & 1 \end{array}$ intersected from inside by the ray $r (t) = (0, 0, 0) + t (1, 0, 0)$ .