Answer to Question #293617 in Analytic Geometry for Elley

Question #293617

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


1
Expert's answer
2022-02-09T13:38:19-0500

For the first find an equation of intersection curve.

{4x2+y2+4z2+4yz+2zx=0x+2y+3z=0\begin{cases} 4x^2+y^2+4z^2+4yz+2zx=0 \\ x+2y+3z=0 \end{cases}



{z=x+2y34x2+y2+4(x+2y3)24y(x+2y)32x(x+2y)3=0\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}



34x2+y28xy=034x^2+y^2-8xy=0

Transform the equation



(y(432i)x)(y(4+32i)x)=0(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 tanα=k2k11+k2k1tan \alpha =\frac{k_{2}-k_{1}}{1+k_{2}k_{1}}


In our case k1=432ik_{1}=4-3\sqrt{2}i and k2=4+32ik_{2}=4+3\sqrt{2}i .

It mean, that tanα=(4+32i)(432i)1+(432i)(4+32i)=62i34=32i17tan\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

α=tan1(32i17)\alpha=tan^{-1}(\frac{3\sqrt{2}i}{17})





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment