Question

Find the points of intersection (if any) of the sphere 2x2 + 2y2 + 272-8x + 4y - 16z + 10 = 0 and the line x= 3 + t, y = 1 + 2t, z=3-t. Explain how you got your answer.

Answer 1

`\left((6+2√(6))/(3), (-3 +4√(6))/(3), (12-2√(6))/(3)\right)\;\;\textsf{and}\;\;\left((6-2√(6))/(3), (-3-4√(6))/(3), (12+2√(6))/(3)\right)`

Explanation:

To find the points of intersection of the sphere and the line, we need to substitute the parametric equations of the line (x = 3 + t, y = 1 + 2t, z = 3 - t) into the equation of the sphere and solve for the parameter t.

The equation of the sphere is given by:

`2x^2 + 2y^2 + 2z^2 - 8x + 4y - 16z + 10 = 0`

Substitute the parametric equations of the line (x = 3 + t, y = 1 + 2t, z = 3 - t) into the equation of the sphere:

`2(3 + t)^2 + 2(1 + 2t)^2 + 2(3 - t)^2 - 8(3 + t) + 4(1 + 2t) - 16(3 - t) + 10 = 0`

Simplify the equation:

`18+12t+2t^2 + 2+8t+8t^2 + 18-12t+2t^2-24-8t +4+8t-48+16t + 10 = 0`

`8t^2+2t^2+2t^2+16t+8t+8t+12t-8t -12t+ 18+ 10+18 +4+ 2-48 -24 = 0`

`12t^2+24t -20 = 0`

Solve for t using the quadratic formula:

`t=(-24 \pm√((24)^2-4(12)(-20)))/(2(12))`

`t=(-24 \pm√(1536))/(24)`

`t=(-24 \pm16√(6))/(24)`

`t=(-3 \pm2√(6))/(3)`

Therefore, the two solutions for t are:

`t=(-3 +2√(6))/(3)`

`t=(-3 -2√(6))/(3)`

Substitute the values of t back into the parametric equations of the line to get the corresponding values of x, y and z:

`\textsf{For}\;\;t=(-3 +2√(6))/(3):`

`x = 3 + \left((-3 +2√(6))/(3)\right)=(6+2√(6))/(3)`

`y = 1 + 2\left((-3 +2√(6))/(3)\right)=(-3 +4√(6))/(3)`

`z = 3 - \left((-3 +2√(6))/(3)\right)=(12-2√(6))/(3)`

Therefore, the point of intersection (x, y, z) is:

`\left((6+2√(6))/(3), (-3 +4√(6))/(3), (12-2√(6))/(3)\right)`

`\textsf{For}\;\;t=(-3-2√(6))/(3):`

`x = 3 + \left((-3-2√(6))/(3)\right)=(6-2√(6))/(3)`

`y = 1 + 2\left((-3-2√(6))/(3)\right)=(-3-4√(6))/(3)`

`z = 3 - \left((-3-2√(6))/(3)\right)=(12+2√(6))/(3)`

Therefore, the point of intersection (x, y, z) is:

`\left((6-2√(6))/(3), (-3-4√(6))/(3), (12+2√(6))/(3)\right)`

So, the two points of intersection of the sphere and line are:

`\left((6+2√(6))/(3), (-3 +4√(6))/(3), (12-2√(6))/(3)\right)\;\;\textsf{and}\;\;\left((6-2√(6))/(3), (-3-4√(6))/(3), (12+2√(6))/(3)\right)`