Question

Derive the equation of the right circular cone generated by rotating the line 6x=3y=2z about the line 2x=-2y=-z

Answer 1

Answer on Question #61283 – Math – Calculus

Question

Derive the equation of the right circular cone generated by rotating the line $6x = 3y = 2z$ about the line $2x = -2y = -z$.

Solution

Write the equation of a line $2x = -2y = -z$ as $\frac{x}{\frac{1}{2}} = \frac{y}{-\frac{1}{2}} = \frac{z}{-1}$. The equation of plane perpendicular to given line has the form $x - y - 2z + D = 0$, where $D$ – constant.

Line represented by equation $\frac{x - x_0}{m} = \frac{y - y_0}{p} = \frac{z - z_0}{q}$ ($\vec{s}(m,p,q)$ is the direction vector of the straight line) and plane $Ax + By + Cz + D = 0$ ($\vec{n}(A,B,C)$ is a normal vector of plane). Hence we get the condition for line and plane $Am + Bp + Cq = 0$ to be parallel. Line is perpendicular to the plane only if the direction vector of the straight collinear to the normal vector of this plane:

$$\frac{A}{m} = \frac{B}{p} = \frac{C}{q}.$$

Plane $x - y - 2z + D = 0$ has normal vector $(1, -1, -2)$, and line $2x = -2y = -z$ or $\frac{x}{\frac{1}{2}} = \frac{y}{-\frac{1}{2}} = \frac{z}{-1}$ has direction vector $\left(\frac{1}{2}, -\frac{1}{2}, -1\right)$. Vectors are collinear, this means that the plane and the straight line are perpendicular.

Find the point of intersection of lines $2x = -2y = -z$ and $6x = 3y = 2z$.

$$\left\{ \begin{array}{l} 6x = 3y = 2z \\ 2x = -2y = -z \end{array} \right. \rightarrow (0,0,0) \text{ – the center of the cone}.$$

Choose some point on the line $2x = -2y = -z$. Let $z = h$, hence $y = \frac{h}{2}$, $x = -\frac{h}{2}$. Find the equation of the plane perpendicular to the axis of rotation passing through the point.

$$-\frac{h}{2} - \frac{h}{2}y - 2h + D = 0 \rightarrow D = 3h \rightarrow x - y - 2z + 3h = 0$$

Find the point of intersection of line $6x = 3y = 2z$ and plane $x - y - 2z + 3h = 0$.

$$\left\{ \begin{array}{l} 6x = 3y = 2z \\ x - y - 2z + 3h = 0 \end{array} \right.$$

Let $z = 6t$, where $t$ some number. From first equation get $t = 4y$, $x = 2t$. Substituting the obtained values into the second equation of the system get

$$2t - 4t - 12t + 3h = 0 \rightarrow 3h = 14t \quad t = \frac{3h}{14}. \text{ Therefore } x = \frac{3h}{14}, y = \frac{6h}{7}, z = \frac{9h}{7}.$$

Find the radius of rotation for that point around the axis $2x = -2y = -z$ using distance formula between two points $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

$$R = \sqrt{\left(\frac{3h}{7} + \frac{h}{2}\right)^2 + \left(\frac{6h}{7} - \frac{h}{2}\right)^2 + \left(\frac{9h}{7} - h\right)^2} = \sqrt{\frac{210}{196}} h.$$

Write the equation of the circle of rotation for the point $x = \frac{3h}{14}, y = \frac{6h}{7}, z = \frac{9h}{7}$.

$$\left\{ \begin{array}{c} x - y - 2 z + 3 h = 0 \\ \left(x + \frac {h}{2}\right) ^ {2} + \left(y - \frac {h}{2}\right) ^ {2} + (z - h) ^ {2} = \frac {2 1 0}{1 9 6} h ^ {2} \end{array} \right.$$

We find from the first equation $h$, and substitute into the second equation and get:

$$\begin{array}{l} h = \frac {y + 2 z - x}{3} \\ \left(x + \frac {y + 2 z - x}{6}\right) ^ {2} + \left(y - \frac {y + 2 z - x}{6}\right) ^ {2} + \left(z - \frac {y + 2 z - x}{3}\right) ^ {2} = \frac {2 1 0}{1 9 6} \left(\frac {y + 2 z - x}{3}\right) ^ {2} \\ \end{array}$$ $$\left(\frac {y + 2 z + 5 x}{6}\right) ^ {2} + \left(\frac {5 y - 2 z + x}{6}\right) ^ {2} + \left(\frac {z - y + x}{3}\right) ^ {2} = \frac {5 (y + 2 z - x) ^ {2}}{4 2}$$

As result get

$$\begin{array}{l} 7 (y + 2 z + 5 x) ^ {2} + 7 (5 y - 2 z + x) ^ {2} + 2 8 (z - y + x) ^ {2} = 3 0 (y + 2 z - x) ^ {2} \\ 7 \left(y ^ {2} + 4 z ^ {2} + 2 5 x ^ {2} + 4 y z + 1 0 x y + 2 0 z x\right) + 7 \left(2 5 y ^ {2} + 4 z ^ {2} + x ^ {2} - 2 0 y z + 1 0 x y - 4 x z\right) + \\ + 2 8 \left(z ^ {2} + y ^ {2} + x ^ {2} - 2 z y - 2 x y + 2 x z\right) = 3 0 \left(4 z ^ {2} + y ^ {2} + x ^ {2} + 4 z y - 2 x y - 4 x z\right) \\ 1 8 0 y ^ {2} + 1 8 0 x ^ {2} - 3 6 z ^ {2} - 2 8 8 z y + 1 4 4 x y + 2 8 8 x z = 0 \\ \end{array}$$

Thus, equation of cone is

$$5 y ^ {2} + 5 x ^ {2} - z ^ {2} - 8 z y + 4 x y + 8 x z = 0$$

Answer: $5y^{2} + 5x^{2} - z^{2} - 8zy + 4xy + 8xz = 0.$

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