Math Problem Statement

Suppose that you stand at the point (2,3,0) and look in the direction of a point that is not visible because it is behind the ball. You will then be looking at a point on the sphere.

If (−4,−5,3) is not visible from (2,3,0), find the point on the sphere at which you are looking if you look in the direction of (−4,−5,3). Otherwise, find the point on the sphere at which you look if you are looking in the direction of (−4,−5,2). Point (x,y,z)=

Equation of sphere is x²+y²+z²=sqrt(13) and please give me answer round to 4 decimal places

Solution

To solve this problem, we first need to determine the equation of the line that passes through the point (2,3,0)(2, 3, 0) in the direction of (4,5,2)(-4, -5, 2), and then find where this line intersects the sphere.

Step 1: Line Equation

The parametric equation of the line can be written as:

(x,y,z)=(2,3,0)+t((42),(53),(20))(x, y, z) = (2, 3, 0) + t((-4 - 2), (-5 - 3), (2 - 0)) (x,y,z)=(2,3,0)+t(6,8,2)(x, y, z) = (2, 3, 0) + t(-6, -8, 2) So, the parametric equations of the line are:

x=26tx = 2 - 6t y=38ty = 3 - 8t z=0+2t=2tz = 0 + 2t = 2t

Step 2: Equation of the Sphere

The equation of the sphere is given as:

x2+y2+z2=13x^2 + y^2 + z^2 = 13

Step 3: Substitution

Now, substitute the parametric equations of the line into the equation of the sphere:

(26t)2+(38t)2+(2t)2=13(2 - 6t)^2 + (3 - 8t)^2 + (2t)^2 = 13

Step 4: Expand and Simplify

Let's expand and simplify the equation step by step.

  1. (26t)2=424t+36t2(2 - 6t)^2 = 4 - 24t + 36t^2
  2. (38t)2=948t+64t2(3 - 8t)^2 = 9 - 48t + 64t^2
  3. (2t)2=4t2(2t)^2 = 4t^2

Now combine them:

424t+36t2+948t+64t2+4t2=134 - 24t + 36t^2 + 9 - 48t + 64t^2 + 4t^2 = 13 4+9+36t2+64t2+4t224t48t=134 + 9 + 36t^2 + 64t^2 + 4t^2 - 24t - 48t = 13 13+104t272t=1313 + 104t^2 - 72t = 13 Simplify further:

104t272t=0104t^2 - 72t = 0

Step 5: Solve the Quadratic Equation

Factor out tt:

t(104t72)=0t(104t - 72) = 0

So, either t=0t = 0 or 104t=72104t = 72, which gives:

t=72104=1826=913t = \frac{72}{104} = \frac{18}{26} = \frac{9}{13}

Step 6: Find the Coordinates

Now substitute t=913t = \frac{9}{13} into the parametric equations for xx, yy, and zz:

x=26(913)=25413=26135413=28132.1538x = 2 - 6\left(\frac{9}{13}\right) = 2 - \frac{54}{13} = \frac{26}{13} - \frac{54}{13} = \frac{-28}{13} \approx -2.1538 y=38(913)=37213=39137213=33132.5385y = 3 - 8\left(\frac{9}{13}\right) = 3 - \frac{72}{13} = \frac{39}{13} - \frac{72}{13} = \frac{-33}{13} \approx -2.5385 z=2(913)=18131.3846z = 2\left(\frac{9}{13}\right) = \frac{18}{13} \approx 1.3846

Final Answer:

The point on the sphere is approximately:

(x,y,z)=(2.1538,2.5385,1.3846)(x, y, z) = (-2.1538, -2.5385, 1.3846)

Would you like more details or have any questions?

Here are 5 related questions:

  1. How do you find the parametric equation of a line in 3D space?
  2. What happens if the point (4,5,3)(-4, -5, 3) is visible from (2,3,0)(2, 3, 0)?
  3. How can you derive the equation of a sphere from its geometric properties?
  4. What are the steps to solving quadratic equations that result from intersections of lines and spheres?
  5. How can parametric equations be used to model motion in 3D space?

Tip: Always simplify parametric equations before substituting into other expressions to minimize calculation errors!

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Parametric Equations
Intersection of Lines and Spheres
Algebra

Formulas

Parametric equation of a line: (x, y, z) = (x0, y0, z0) + t(dx, dy, dz)
Equation of a sphere: x^2 + y^2 + z^2 = r^2
Quadratic equation: ax^2 + bx + c = 0

Theorems

Pythagorean Theorem
Quadratic Formula

Suitable Grade Level

Grades 10-12 (High School Geometry and Algebra)