Answer on Question #46071 – Math – Analytic Geometry
Problem.
Find the path traced by the centre of the sphere which touches the lines
x / y = y / ( − 1 ) = z / 2 and 2 x = y , y − z = 0. x/y = y/(-1) = z/2 \text{ and } 2x = y, y - z = 0. x / y = y / ( − 1 ) = z /2 and 2 x = y , y − z = 0.
Solution:
We will rewrite lines x y = y − 1 = z 2 \frac{x}{y} = \frac{y}{-1} = \frac{z}{2} y x = − 1 y = 2 z 2 x = y , y − z = 0 2x = y, y - z = 0 2 x = y , y − z = 0
Let t ∈ R t \in \mathbb{R} t ∈ R x y = y − 1 = z 2 = t \frac{x}{y} = \frac{y}{-1} = \frac{z}{2} = t y x = − 1 y = 2 z = t x = − t 2 x = -t^2 x = − t 2 y = − t y = -t y = − t z = 2 t z = 2t z = 2 t
Hence x y = y − 1 = z 2 = t \frac{x}{y} = \frac{y}{-1} = \frac{z}{2} = t y x = − 1 y = 2 z = t
Let s ∈ R s \in \mathbb{R} s ∈ R x = s x = s x = s x = s x = s x = s y = 2 s y = 2s y = 2 s z = 2 s z = 2s z = 2 s
The tangent vector of the first ( − 2 t , − 1 , 2 ) (-2t, -1, 2) ( − 2 t , − 1 , 2 )
The direction vectors of the second line is ( 1 , 2 , 2 ) (1,2,2) ( 1 , 2 , 2 )
Suppose that ( x 0 , y 0 , z 0 ) (x_0,y_0,z_0) ( x 0 , y 0 , z 0 )
( x 0 − t 2 ) 2 + ( y 0 + t ) 2 + ( z 0 − 2 t ) 2 = ( x 0 − s ) 2 + ( y 0 − 2 s ) 2 + ( z 0 − 2 s ) 2 , \sqrt{(x_0 - t^2)^2 + (y_0 + t)^2 + (z_0 - 2t)^2} = \sqrt{(x_0 - s)^2 + (y_0 - 2s)^2 + (z_0 - 2s)^2}, ( x 0 − t 2 ) 2 + ( y 0 + t ) 2 + ( z 0 − 2 t ) 2 = ( x 0 − s ) 2 + ( y 0 − 2 s ) 2 + ( z 0 − 2 s ) 2 ,
as the radiuses of the sphere.
If
( x 0 − t 2 ) 2 + ( y 0 + t ) 2 + ( z 0 − 2 t ) 2 = ( x 0 − s ) 2 + ( y 0 − 2 s ) 2 + ( z 0 − 2 s ) 2 , \sqrt{(x_0 - t^2)^2 + (y_0 + t)^2 + (z_0 - 2t)^2} = \sqrt{(x_0 - s)^2 + (y_0 - 2s)^2 + (z_0 - 2s)^2}, ( x 0 − t 2 ) 2 + ( y 0 + t ) 2 + ( z 0 − 2 t ) 2 = ( x 0 − s ) 2 + ( y 0 − 2 s ) 2 + ( z 0 − 2 s ) 2 ,
then
x 0 2 − 2 x 0 t 2 + t 4 + y 0 2 + 2 y 0 t + t 2 + z 0 − 4 z 0 t + 4 t 2 = x 0 − 2 x 0 s + s 2 + y 0 2 − 4 y 0 s + 4 s 2 + z 0 2 − 4 z 0 s + 4 s 2 \begin{array}{l}
x_0^2 - 2x_0t^2 + t^4 + y_0^2 + 2y_0t + t^2 + z_0 - 4z_0t + 4t^2 \\
= x_0 - 2x_0s + s^2 + y_0^2 - 4y_0s + 4s^2 + z_0^2 - 4z_0s + 4s^2
\end{array} x 0 2 − 2 x 0 t 2 + t 4 + y 0 2 + 2 y 0 t + t 2 + z 0 − 4 z 0 t + 4 t 2 = x 0 − 2 x 0 s + s 2 + y 0 2 − 4 y 0 s + 4 s 2 + z 0 2 − 4 z 0 s + 4 s 2
or
− 2 t ( x 0 t − y 0 + 2 z 0 ) + t 4 + 5 t 2 = − 2 s ( x 0 + 2 y 0 + 2 z 0 ) + 9 s 2 . -2t(x_0t - y_0 + 2z_0) + t^4 + 5t^2 = -2s(x_0 + 2y_0 + 2z_0) + 9s^2. − 2 t ( x 0 t − y 0 + 2 z 0 ) + t 4 + 5 t 2 = − 2 s ( x 0 + 2 y 0 + 2 z 0 ) + 9 s 2 .
The sphere should touches the lines (the direction vector of the curves should be perpendicular to the vector from the center of the sphere to the point on the line), so
( x 0 + t 2 ) ⋅ ( − 2 t ) + ( y 0 + t ) ⋅ ( − 1 ) + ( z 0 − 2 t ) ⋅ 2 = 0 (x_0 + t^2) \cdot (-2t) + (y_0 + t) \cdot (-1) + (z_0 - 2t) \cdot 2 = 0 ( x 0 + t 2 ) ⋅ ( − 2 t ) + ( y 0 + t ) ⋅ ( − 1 ) + ( z 0 − 2 t ) ⋅ 2 = 0
and
( x 0 − s ) ⋅ 1 + ( y 0 − 2 s ) ⋅ 2 + ( z 0 − 2 s ) ⋅ 2 = 0. (x_0 - s) \cdot 1 + (y_0 - 2s) \cdot 2 + (z_0 - 2s) \cdot 2 = 0. ( x 0 − s ) ⋅ 1 + ( y 0 − 2 s ) ⋅ 2 + ( z 0 − 2 s ) ⋅ 2 = 0.
Therefore
− 2 t x 0 − y 0 + 2 z 0 = 2 t 3 + 5 t or − y 0 + 2 z 0 = 2 t 3 + 5 t + 2 t x 0 -2tx_0 - y_0 + 2z_0 = 2t^3 + 5t \text{ or } -y_0 + 2z_0 = 2t^3 + 5t + 2tx_0 − 2 t x 0 − y 0 + 2 z 0 = 2 t 3 + 5 t or − y 0 + 2 z 0 = 2 t 3 + 5 t + 2 t x 0
and
x 0 + 2 y 0 + 2 z 0 = 9 s . x_0 + 2y_0 + 2z_0 = 9s. x 0 + 2 y 0 + 2 z 0 = 9 s .
Therefore
− 2 t ( x 0 t + 2 t 3 + 5 t + 2 t x 0 ) + t 4 + 5 t 2 = − 18 s 2 + 9 s 2 -2t(x_0t + 2t^3 + 5t + 2tx_0) + t^4 + 5t^2 = -18s^2 + 9s^2 − 2 t ( x 0 t + 2 t 3 + 5 t + 2 t x 0 ) + t 4 + 5 t 2 = − 18 s 2 + 9 s 2
or
− 6 t x 0 − 3 t 4 − 5 t 2 = − 9 s 2 . -6tx_0 - 3t^4 - 5t^2 = -9s^2. − 6 t x 0 − 3 t 4 − 5 t 2 = − 9 s 2 .
Hence
x 0 = 9 s 2 − 3 t 4 − 5 t 2 6 t , 2 y 0 + 2 z 0 = 9 s − 9 s 2 − 3 t 4 − 5 t 2 6 t , − y 0 + 2 z 0 = 2 t 3 + 5 t + 9 s 2 − 3 t 4 − 5 t 2 3 \begin{array}{l}
x_0 = \frac{9s^2 - 3t^4 - 5t^2}{6t}, \quad 2y_0 + 2z_0 = 9s - \frac{9s^2 - 3t^4 - 5t^2}{6t}, \\
\quad -y_0 + 2z_0 = 2t^3 + 5t + \frac{9s^2 - 3t^4 - 5t^2}{3}
\end{array} x 0 = 6 t 9 s 2 − 3 t 4 − 5 t 2 , 2 y 0 + 2 z 0 = 9 s − 6 t 9 s 2 − 3 t 4 − 5 t 2 , − y 0 + 2 z 0 = 2 t 3 + 5 t + 3 9 s 2 − 3 t 4 − 5 t 2
Then y 0 = 9 s + 2 t 3 + 5 t − 9 s 2 − 3 t 4 − 5 t 2 6 t + 9 s 2 − 3 t 4 − 5 t 2 3 y_0 = 9s + 2t^3 + 5t - \frac{9s^2 - 3t^4 - 5t^2}{6t} + \frac{9s^2 - 3t^4 - 5t^2}{3} y 0 = 9 s + 2 t 3 + 5 t − 6 t 9 s 2 − 3 t 4 − 5 t 2 + 3 9 s 2 − 3 t 4 − 5 t 2 z 0 = 2 t 3 + 5 t + 9 2 s − 9 s 2 − 3 t 4 − 5 t 2 12 t + 9 s 2 − 3 t 4 − 5 t 2 3 z_0 = 2t^3 + 5t + \frac{9}{2}s - \frac{9s^2 - 3t^4 - 5t^2}{12t} + \frac{9s^2 - 3t^4 - 5t^2}{3} z 0 = 2 t 3 + 5 t + 2 9 s − 12 t 9 s 2 − 3 t 4 − 5 t 2 + 3 9 s 2 − 3 t 4 − 5 t 2
Answer:
x 0 = 9 s 2 − 3 t 4 − 5 t 2 6 t , y 0 = 9 s + 2 t 3 + 5 t − 9 s 2 − 3 t 4 − 5 t 2 6 t + 9 s 2 − 3 t 4 − 5 t 2 3 , z 0 = 2 t 3 + 5 t + 9 2 s − 9 s 2 − 3 t 4 − 5 t 2 12 t + 9 s 2 − 3 t 4 − 5 t 2 3 . \begin{array}{l}
x_0 = \frac{9s^2 - 3t^4 - 5t^2}{6t}, \quad y_0 = 9s + 2t^3 + 5t - \frac{9s^2 - 3t^4 - 5t^2}{6t} + \frac{9s^2 - 3t^4 - 5t^2}{3}, \quad z_0 = 2t^3 + 5t + \frac{9}{2}s - \\
\frac{9s^2 - 3t^4 - 5t^2}{12t} + \frac{9s^2 - 3t^4 - 5t^2}{3}.
\end{array} x 0 = 6 t 9 s 2 − 3 t 4 − 5 t 2 , y 0 = 9 s + 2 t 3 + 5 t − 6 t 9 s 2 − 3 t 4 − 5 t 2 + 3 9 s 2 − 3 t 4 − 5 t 2 , z 0 = 2 t 3 + 5 t + 2 9 s − 12 t 9 s 2 − 3 t 4 − 5 t 2 + 3 9 s 2 − 3 t 4 − 5 t 2 .
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#340153 on Dec 2023