Question

Find the angle between the lines of intersection of the cone 4𝑥² + 𝑦² + 4𝑧² + 4𝑦𝑧 + 2𝑧𝑥 = 0 and the plane 𝑥 + 2𝑦 + 3𝑧 = 0.

Answer 1

For the first find an equation of intersection curve.

$$\begin{cases} 4x^2+y^2+4z^2+4yz+2zx=0 \\ x+2y+3z=0 \end{cases}$$

$$\begin{cases} z=-\frac{x+2y}{3} \\ 4x^2+y^2+4(\frac{x+2y}{3})^2-\frac{4y(x+2y)}{3}-\frac{2x(x+2y)}{3}=0 \end{cases}$$

$$34x^2+y^2-8xy=0$$

Transform the equation

$$(y-(4-3\sqrt{2}i)x)(y-(4+3\sqrt{2}i)x)=0$$

This is equation of two imaginary intersecting lines.

Angle between the lines y=k₁x and y=k₂x is $tan \alpha =\frac{k_{2}-k_{1}}{1+k_{2}k_{1}}$

In our case $k_{1}=4-3\sqrt{2}i$ and $k_{2}=4+3\sqrt{2}i$ .

It mean, that $tan\alpha =\frac{(4+3\sqrt{2}i)-(4-3\sqrt{2}i)}{1+(4-3\sqrt{2}i)(4+3\sqrt{2}i)}=\frac{6\sqrt{2}i}{34}=\frac{3\sqrt{2}i}{17}$

The angle between the lines of intersection is

$$\alpha=tan^{-1}(\frac{3\sqrt{2}i}{17})$$