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Chapter 8: Problem 13

A line with parametric equations \(x=10+2 s, y=5+s, z=2, s \in \mathbf{R}\), intersects a sphere with the equation \(x^{2}+y^{2}+z^{2}=9\) at the points \(A\) and \(B\). Determine the coordinates of these points.

Short Answer

Expert verified
The points of intersection are \(A(2, 1, 2)\) and \(B(-2, -1, 2)\).

Step by step solution

01

Substitute Parametric Equations into Sphere Equation

First, we substitute the parametric equations into the sphere's equation. The sphere's equation is:\[x^2 + y^2 + z^2 = 9\]Substitute the parametric equations:\[(10+2s)^2 + (5+s)^2 + 2^2 = 9\]
02

Expand and Simplify the Equation

Expand the equation:\[(100 + 40s + 4s^2) + (25 + 10s + s^2) + 4 = 9\]Combine like terms:\[4s^2 + s^2 + 50s + 129 = 9\]This simplifies to:\[5s^2 + 50s + 120 = 0\]
03

Set Equation to Zero

Subtract 9 from both sides to set the equation to zero:\[5s^2 + 50s + 120 = 0\]Simplify by dividing through by 5:\[s^2 + 10s + 24 = 0\]
04

Solve the Quadratic Equation

Use the quadratic formula, \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=10\), and \(c=24\).Calculate:\[s = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times 24}}{2 \times 1}\]\[s = \frac{-10 \pm \sqrt{100 - 96}}{2}\]\[s = \frac{-10 \pm 2}{2}\]This gives: \(s_1 = -4, s_2 = -6\).
05

Find Coordinates of Intersection Points A and B

Substitute \(s = -4\) back into the parametric equations:\[A: (x, y, z) = (10 + 2(-4), 5 - 4, 2) = (2, 1, 2)\]Substitute \(s = -6\) back into the equations:\[B: (x, y, z) = (10 + 2(-6), 5 - 6, 2) = (-2, -1, 2)\]
06

Final Step: Conclusion

The coordinates of the points where the line intersects the sphere are \(A(2, 1, 2)\) and \(B(-2, -1, 2)\).

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Key Concepts

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

Parametric Equations
Parametric equations are a powerful tool in vector calculus. They allow us to describe a line or curve in space using a set of equations. Each equation represents a coordinate in the space.
  • In our example, the line is described by the parametric equations: \( x = 10 + 2s, y = 5 + s, z = 2 \).
  • Here, \( s \) is a parameter that can take any real number value, meaning as \( s \) changes, it traces out the line in 3D space.
  • The coefficients in these equations, such as 10, 5, or 2, and the terms involving \( s \), \( 2s \) or \( s \), determine the direction and position of the line.
By substituting different values of \( s \), we can find any number of points along this line.
Sphere Intersection
The sphere equation gives us a geometric shape in which every point is equidistant from a central point. The equation for a sphere, centered at the origin, is \( x^2 + y^2 + z^2 = r^2 \), where \( r \) is the radius.
  • In our problem, the sphere equation is \( x^2 + y^2 + z^2 = 9 \), where 9 is \( r^2 \), indicating a sphere with a radius of 3.
  • To find where the line intersects the sphere, we substitute the line's parametric equations into the sphere equation.
By doing this, we develop an equation in terms of \( s \) that tells us for which values the line enters and exits the sphere.
Quadratic Formula
The quadratic formula is a savior when it comes to solving quadratic equations, typically those in the form \( ax^2 + bx + c = 0 \). Given our equation from the sphere intersection, we found \( 5s^2 + 50s + 120 = 0 \) and simplified it to\( s^2 + 10s + 24 = 0 \).
  • To find the values of \( s \) (where the line intersects the sphere), we use the quadratic formula: \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • By inserting our coefficients \( a = 1 \), \( b = 10 \), \( c = 24 \), we solve for \( s \).
  • This formula provides the potential values of \( s \) at which intersection occurs, simplifying and solving the complex equation.
This results in the values \( s_1 = -4 \) and \( s_2 = -6 \), indicating precisely where the intersections are.
Coordinates of Points
Once we have solved for \( s \), finding the coordinates for the intersection points becomes straightforward by substituting \( s_1 = -4 \) and \( s_2 = -6 \) back into the parametric equations.
  • For \( s = -4 \), substituting into the parametric equations gives the point \( A: (x, y, z) = (10 + 2(-4), 5 - 4, 2) = (2, 1, 2) \).
  • Similarly, for \( s = -6 \), substituting gives the point \( B: (x, y, z) = (10 + 2(-6), 5 - 6, 2) = (-2, -1, 2) \).
These coordinates \( A(2, 1, 2) \) and \( B(-2, -1, 2) \) are the exact points where the line intersects the sphere.

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Most popular questions from this chapter

A plane has \(\vec{r}=(-3,5,6)+s(-1,1,2)+v(2,1,-3), s, v \in \mathbf{R}\) as its equation. a. Give the equations of two intersecting lines that lie on this plane. b. What point do these two lines have in common?

The plane with equation \(\vec{r}=(1,2,3)+m(1,2,5)+n(1,-1,3)\) intersects the \(y\) - and \(z\) -axes at the points \(A\) and \(B\), respectively. Determine the equation of the line that contains these two points.

Determine the equation of the plane that contains the point \(P(-1,2,1)\) and the line \(\vec{r}=(2,1,3)+s(4,1,5), s \in \mathbf{R}.\)

For the line \(L: \vec{r}=(1,-5)+s(3,5), s \in \mathbf{R},\) determine the following: a. an equation for the line perpendicular to \(L,\) passing through \(P(2,0)\) b. the point at which the line in part a. intersects the \(y\) -axis

a. Determine parameters corresponding to the point \(P(5,3,2),\) where \(P\) is a point on the plane with equation \(\pi: \vec{r}=(2,0,1)+s(4,2,-1)+t(-1,1,2), s, t \in \mathbf{R}\) b. Show that the point \(A(0,5,-4)\) does not lie on \(\pi\)
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