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

$$1) x - y + 2z = 0 \leftrightarrow z = \frac{(y-x)}{2}$$ $$2)x^2 +y^2 -4z^2 +6yz = 0$$

After placing z out 1) to 2) we get

$$3)x^2 + y^2 - (y-x)^2 + 3y(y-x) = 0 \leftrightarrow y(3y-x)=0$$

Let's find intersection line equations out of 3)

$$a) x = t, y = \frac{1}{3}t, z = -\frac{1}{3}t$$ $$b) x = t, y = 0, z = - \frac{1}{2}t$$

then we find points on this lines and build collinear vectors of them

$$a)A(3, 1, -1), B(6, 2,-2), \overline{AB} = (3, 1, -1)$$

$$b)C(2, 0, -1), D(4, 0,-2), \overline{CD} = (2, 0, -1)$$

Then we use the scalar product formula to find the angle between vectors lying on intersection lines

$$\overline{AB}\cdot \overline{CD} = |\overline{AB}| |\overline{CD}|cos(\angle(\overline{AB} , \overline{CD}) )$$

$$7 = \sqrt{5}\sqrt{11}cos(\angle(\overline{AB} , \overline{CD}) )$$

$$cos{\angle(\overline{AB} , \overline{ CD}) } = \sqrt{\frac{49}{55}}\implies\angle(\overline{AB} , \overline{CD}) \approx 0.336.$$

$$\arctan{\angle(\overline{AB} , \overline{CD}) } = \frac{\pi}{2} - \arccos(\tfrac{\angle(\overline{AB} , \overline{CD}) }{\sqrt{1 +\angle(\overline{AB} , \overline{CD})^2 }}) \approx$$

$$\approx 1.57 - 1.24 = 0.33 \ne 0.746 \approx \arctan(\sqrt{6/7})$$