Question

Show that the angle between the two lines in which the plane x−y+2z = 0 intersects the cone

x² +y² −4z² +6yz = 0 is tan ⁻¹ (√ 6 /7)

Answer 1

Answer:-

The lines of intersection have equation

$\begin{cases} x^2+y^2-4z^2+6yz=0; \\ x-y+2z=0; \\ z=t, \end{cases}$

where 𝑡 ∈ ℝ.

The system  $\begin{cases} x^2+y^2-4z^2+6yz=0; \\ x-y+2z=0; \\ z=t, \end{cases}$

is equivalent to $\begin{cases} x^2+y^2-4t^2+6yt=0; \\ x=y-2t; \\ z=t, \end{cases}$

or $\begin{cases} y^3-4yt+4t^2+y^2-4t^2+6yt=0; \\ x=y-2t; \\ z=t, \end{cases}$

or $\begin{cases} y(y+t)=0; \\ x=y-2t; \\ z=t, \end{cases}$

we obtain two line

$\begin{cases} x=-2t; \\ y=0; \\ z=t, \end{cases}$ and $\begin{cases} x=-3t; \\ y=-t; \\ z=t, \end{cases}$

where 𝑡 is real parameter. The direction vector of the first line is 𝑣₁ = (−2,0,1) and the direction vector of the second line is 𝑣₂ = (−3, −1,1). The angle between the two lines is equal to the angle between the direction vectors of this two lines.

cos(𝑣̂₁, 𝑣₂ ) =$\frac{v_1.v_2}{|v_1||v_2|}=\frac{6+1}{\sqrt{5}\sqrt{11}}=\sqrt{\frac{49}{55}}$

tan²(𝑣̂₁, 𝑣₂ ) =$\frac{1}{cos^2(v_1.v_2)}-1=\frac{55}{49}-1=\frac{6}{49}$

The (𝑣̂₁, 𝑣₂ ) = arctan$√6 \over7$ , as cos(𝑣̂₁, 𝑣₂ ) is positive. Therefore, the angle between two lines is equal to arctan $√6\over 7$ =  tan ⁻¹ ($√ 6 \over7$ )