Intersection of a plane and cone

Question

Find the angle between the lines in which the cone $$4x^2-y^2+3z^2=0$$
is cut by the plane $$2x+y-z=0$$.

My solution- since a plane cuts the cone in infinite number of lines, which are generators, i need to find angle between which two of them ??

obviously by taking a line $$x/l=y/m=z/n$$ and solving it with given two equations , i will get two lines whose angle of intersection is easy to find out but i don't see how this makes sense .

I guess you mean the cone $4x^2-y^2+3z^2=0$ and the plane $2x+y-z=0$? — Matthew Leingang, May 01 '20 at 18:59
You have a serious problem! $4x^2- y^2+ z^2$ is not a cone and $2x+ y- z$ is not a plane! Did you intend for the "-" signs to be "=" signs? — user247327, May 01 '20 at 19:00

score 1 · Accepted Answer · answered May 01 '20 at 19:55

From $4x^2-y^2+3z^2=0$ and $2x+y-z=0$, we have

$$2z(z+2x)=0$$

which are the $xz$-projections of the two intersecting lines. Then, $z=0$ and $z+2x=0$ lead to the equations of the lines below

$$L_1=t(1,-2,0),\>\>\>\>\>L_2= s(1,-4,-2)$$

ang the angle $\theta$ between them is given by

$$\cos\theta = \pm \frac{(1,-2,0)\cdot(1,-4,-2)}{ \sqrt{5}\cdot\sqrt{21} }= \pm\frac9{\sqrt {105}}$$

Note that there are two ways to measure the angle.

score 0 · Answer 2 · answered May 01 '20 at 18:59

0

I got it, i was visualizing it in the wrong way.
The plane cuts the cone at vertex and parallel to axis.
Proceeding with the method described in my question, we get
cos$\theta$=-9/$\sqrt105$

answered May 01 '20 at 18:59

George carlin

558

Intersection of a plane and cone

2 Answers2