Question

Find the distance of the centre of the circle x^2+y^2+z^2+x-2y+2z=3, 2x+y+2z=1 from the plane ax+by+cz=d where a,b,c,d are constants and also find the equation of the right circular cylinder whose base curve is the circle obtained above.
above.

Answer 1

Answer on Question #78894 - Math - Analytic Geometry

Find the distance of the centre of the circle $x^{2} + y^{2} + z^{2} + x - 2y + 2z = 3$, $2x + y + 2z = 1$ from the plane $ax + by + cz = d$ where $a, b, c, d$ are constants and also find the equation of the right circular cylinder whose base curve is the circle obtained above.

Solution:

1. $x^{2} + y^{2} + z^{2} + x - 2y + 2z = 3$

$$x ^ {2} + 2 \cdot 0. 5 \cdot x + 0. 2 5 + y ^ {2} - 2 y + 1 + z ^ {2} + 2 z + 1 - 2. 2 5 = 3$$ $$(x + 0. 5) ^ {2} + (y - 1) ^ {2} + (z + 1) ^ {2} = \frac {2 1}{4}$$

$x^{2} + y^{2} + z^{2} + x - 2y + 2z = 3$ is an equation of the sphere with centre in $r_0 = (-0.5,1, - 1)$ and $R = \frac{\sqrt{21}}{2}$.

2. For plane $2x + y + 2z = 1$ normal vector is $v = (2,1,2)$.

3. Centre of the circle $x^{2} + y^{2} + z^{2} + x - 2y + 2z = 3$, $2x + y + 2z = 1$ lies on radius perpendicular to the plane and can be found from following system:

$$\left\{ \begin{array}{l} x _ {0} = - 0. 5 + 2 \cdot \lambda \\ y _ {0} = 1 + 1 \cdot \lambda \\ z _ {0} = - 1 + 2 \cdot \lambda \\ 2 x _ {0} + y _ {0} + 2 z _ {0} = 1 \end{array} \right.$$ $$\left\{ \begin{array}{l} \lambda = \frac {1}{3} \\ x _ {0} = \frac {1}{6} \\ y _ {0} = \frac {8}{6} \\ z _ {0} = - \frac {2}{6} \end{array} \right.$$

Distance between centre of sphere and centre of circle: $h = \lambda \cdot \sqrt{2^2 + 1^2 + 2^2} = 1$

4. Distance of the centre of the circle from the plane $ax + by + cz = d$ can be found from following expression:

$$d = \frac {| a x _ {0} + b y _ {0} + c z _ {0} - d |}{\sqrt {a ^ {2} + b ^ {2} + c ^ {2}}} = \frac {| a + 8 b - 2 c - 6 d |}{6 \sqrt {a ^ {2} + b ^ {2} + c ^ {2}}}$$

5. Radius of circle: $R_{c} = \sqrt{R^{2} - h^{2}} = \sqrt{\frac{21}{4} - 1} = \frac{\sqrt{17}}{2}$

6. Two orthonormal vectors on plane $2x + y + 2z = 1$: $k = \left(-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}\right)$, $l = \left(\frac{\sqrt{2}}{6}, - \frac{2\sqrt{2}}{3},\frac{\sqrt{2}}{6}\right)$

7. Equation of right cylinder:

$$r (\varphi , \lambda) = r _ {0} + R _ {c} (c o s \varphi \cdot k + s i n \varphi \cdot l) + \lambda \cdot v \qquad \varphi \in [ 0; 2 \pi) \quad \lambda \in (- \infty ; \infty)$$

with

$$r _ {0} = (- 0. 5, 1, - 1)$$ $$R _ {c} = \frac {\sqrt {1 7}}{2}$$ $$k = \left(- \frac {\sqrt {2}}{2}, 0, \frac {\sqrt {2}}{2}\right), \qquad l = \left(\frac {\sqrt {2}}{6}, - \frac {2 \sqrt {2}}{3}, \frac {\sqrt {2}}{6}\right)$$ $$v = (2, 1, 2)$$

Answer:

Distance of the centre of the circle from the plane $d = \frac{|a + 8b - 2c - 6d|}{6\sqrt{a^2 + b^2 + c^2}}$

Equation of cylinder: $r(\varphi, \lambda) = r_0 + R_c(cos\varphi \cdot k + \sin\varphi \cdot l) + \lambda \cdot v$ (with the previously mentioned variables)

Answer provided by https://www.AssignmentExpert.com