Question

Show that the angle between the two lines in which the plane x - y + 2z = 0 intersects the cone x^2 + y^2 - 4z^2 + 6yz = 0 is tan inverse of (under-root 6 / 7) .

Answer 1

Answer on Question #45095 – Math- Analytic Geometry

Problem.

Show that the angle between the two lines in which the plane $x - y + 2z = 0$ intersects the cone $x^2 + y^2 - 4z^2 + 6yz = 0$ is tan inverse of (under-root 6 / 7).

Solution.

The lines of intersection have equation

$$\left\{ \begin{array}{l} x ^ {2} + y ^ {2} - 4 z ^ {2} + 6 y z = 0; \\ x - y + 2 z = 0; \\ z = t, \end{array} \right.$$

where $t\in \mathbb{R}$.

The system

$$\left\{ \begin{array}{l} x ^ {2} + y ^ {2} - 4 z ^ {2} + 6 y z = 0; \\ x - y + 2 z = 0; \\ z = t. \end{array} \right.$$

is equivalent to

$$\left\{ \begin{array}{l} x ^ {2} + y ^ {2} - 4 t ^ {2} + 6 y t = 0; \\ x = y - 2 t; \\ z = t; \end{array} \right.$$

or

$$\left\{ \begin{array}{l} y ^ {2} - 4 y t + 4 t ^ {2} + y ^ {2} - 4 t ^ {2} + 6 y t = 0; \\ x = y - 2 t; \\ z = t; \end{array} \right.$$

or

$$\left\{ \begin{array}{l} y (y + t) = 0; \\ x = y - 2 t; \\ z = t. \end{array} \right.$$

We obtain two lines

$$\left\{ \begin{array}{l} x = - 2 t; \\ y = 0; \\ z = t; \end{array} \right. \quad \text{and} \quad \left\{ \begin{array}{l} x = - 3 t; \\ y = - t; \\ z = t, \end{array} \right.$$

where $t$ is real parameter.

The direction vector of the first line is $v_{1} = (-2,0,1)$ and the direction vector of the second line is $v_{2} = (-3, -1,1)$. The angle between the two lines is equal to the angle between the direction vectors of this two lines.

$$\cos (\widehat {v _ {1} , v _ {2}}) = \frac {(v _ {1} , v _ {2})}{| v _ {1} | \cdot | v _ {2} |} = \frac {6 + 1}{\sqrt {5} \cdot \sqrt {11}} = \sqrt {\frac {49}{55}}.$$ $$\tan^ {2} (\widehat {v _ {1} , v _ {2}}) = \frac {1}{\cos^ {2} (\widehat {v _ {1} , v _ {2}})} - 1 = \frac {55}{49} - 1 = \frac {6}{49}.$$

The $(\widehat{v_1,v_2}) = \arctan \frac{\sqrt{6}}{7}$, as $\cos (\widehat{v_1,v_2})$ is positive.

Therefore, the angle between two lines is equal to $\arctan \frac{\sqrt{6}}{7}$.

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