Mister Exam
r(2i+6j-4k)=3i+j+3k canonical form
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The solution
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-4*I - 3*k + r*(-4*k + 8*I) = 0
$$- 3 k + r \left(- 4 k + 8 i\right) - 4 i = 0$$
-3*k + r*(-4*k + 8*i) - 4*i = 0
Detail solution
Given line equation of 2-order:
$$- 3 k + r \left(- 4 k + 8 i\right) - 4 i = 0$$
This equation looks like:
$$a_{11} r^{2} + 2 a_{12} k r + 2 a_{13} r + a_{22} k^{2} + 2 a_{23} k + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = -2$$
$$a_{13} = 4 i$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3}{2}$$
$$a_{33} = - 4 i$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & -2\\-2 & 0\end{matrix}\right|$$
$$\Delta = -4$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} r_{0} + a_{12} k_{0} + a_{13} = 0$$
$$a_{12} r_{0} + a_{22} k_{0} + a_{23} = 0$$
substitute coefficients
$$- 2 k_{0} + 4 i = 0$$
$$- 2 r_{0} - \frac{3}{2} = 0$$
then
$$r_{0} = - \frac{3}{4}$$
$$k_{0} = 2 i$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} r'^{2} + 2 a_{12} k' r' + a_{22} k'^{2} = 0$$
where
$$a'_{33} = a_{13} r_{0} + a_{23} k_{0} + a_{33}$$
or
$$a'_{33} = - \frac{3 k_{0}}{2} + 4 i r_{0} - 4 i$$
$$a'_{33} = - 10 i$$
then equation turns into
$$- 4 k' r' - 10 i = 0$$
Rotate the resulting coordinate system by an angle φ
$$r' = - \tilde k \sin{\left(\phi \right)} + \tilde r \cos{\left(\phi \right)}$$
$$k' = \tilde k \cos{\left(\phi \right)} + \tilde r \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$r' = - \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}$$
$$k' = \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}$$
then the equation turns from
$$- 4 k' r' - 10 i = 0$$
to
$$- 4 \left(- \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}\right) \left(\frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}\right) - 10 i = 0$$
simplify
$$2 \tilde k^{2} - 2 \tilde r^{2} - 10 i = 0$$
$$- 2 \tilde k^{2} + 2 \tilde r^{2} + 10 i = 0$$
Given equation is hyperbole
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$- 3 k + r \left(- 4 k + 8 i\right) - 4 i = 0$$
This equation looks like:
$$a_{11} r^{2} + 2 a_{12} k r + 2 a_{13} r + a_{22} k^{2} + 2 a_{23} k + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = -2$$
$$a_{13} = 4 i$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3}{2}$$
$$a_{33} = - 4 i$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & -2\\-2 & 0\end{matrix}\right|$$
$$\Delta = -4$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} r_{0} + a_{12} k_{0} + a_{13} = 0$$
$$a_{12} r_{0} + a_{22} k_{0} + a_{23} = 0$$
substitute coefficients
$$- 2 k_{0} + 4 i = 0$$
$$- 2 r_{0} - \frac{3}{2} = 0$$
then
$$r_{0} = - \frac{3}{4}$$
$$k_{0} = 2 i$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} r'^{2} + 2 a_{12} k' r' + a_{22} k'^{2} = 0$$
where
$$a'_{33} = a_{13} r_{0} + a_{23} k_{0} + a_{33}$$
or
$$a'_{33} = - \frac{3 k_{0}}{2} + 4 i r_{0} - 4 i$$
$$a'_{33} = - 10 i$$
then equation turns into
$$- 4 k' r' - 10 i = 0$$
Rotate the resulting coordinate system by an angle φ
$$r' = - \tilde k \sin{\left(\phi \right)} + \tilde r \cos{\left(\phi \right)}$$
$$k' = \tilde k \cos{\left(\phi \right)} + \tilde r \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$r' = - \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}$$
$$k' = \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}$$
then the equation turns from
$$- 4 k' r' - 10 i = 0$$
to
$$- 4 \left(- \frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}\right) \left(\frac{\sqrt{2} \tilde k}{2} + \frac{\sqrt{2} \tilde r}{2}\right) - 10 i = 0$$
simplify
$$2 \tilde k^{2} - 2 \tilde r^{2} - 10 i = 0$$
$$- 2 \tilde k^{2} + 2 \tilde r^{2} + 10 i = 0$$
Given equation is hyperbole
None
- reduced to canonical form
The center of canonical coordinate system at point O
(-3/4, 2*I)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$- 3 k + r \left(- 4 k + 8 i\right) - 4 i = 0$$
This equation looks like:
$$a_{11} r^{2} + 2 a_{12} k r + 2 a_{13} r + a_{22} k^{2} + 2 a_{23} k + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = -2$$
$$a_{13} = 4 i$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3}{2}$$
$$a_{33} = - 4 i$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & -2 & 4 i\\-2 & 0 & - \frac{3}{2}\\4 i & - \frac{3}{2} & - 4 i\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & -2\\-2 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -4$$
$$I_{3} = 40 i$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4$$
$$K_{2} = \frac{55}{4}$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 4 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = 2$$
then the canonical form of the equation will be
$$\tilde k^{2} \lambda_{2} + \tilde r^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$2 \tilde k^{2} - 2 \tilde r^{2} - 10 i = 0$$
- reduced to canonical form
$$- 3 k + r \left(- 4 k + 8 i\right) - 4 i = 0$$
This equation looks like:
$$a_{11} r^{2} + 2 a_{12} k r + 2 a_{13} r + a_{22} k^{2} + 2 a_{23} k + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = -2$$
$$a_{13} = 4 i$$
$$a_{22} = 0$$
$$a_{23} = - \frac{3}{2}$$
$$a_{33} = - 4 i$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = 0$$
|0 -2|
I2 = | |
|-2 0 |$$I_{3} = \left|\begin{matrix}0 & -2 & 4 i\\-2 & 0 & - \frac{3}{2}\\4 i & - \frac{3}{2} & - 4 i\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & -2\\-2 & - \lambda\end{matrix}\right|$$
| 0 4*I | | 0 -3/2|
K2 = | | + | |
|4*I -4*I| |-3/2 -4*I|$$I_{1} = 0$$
$$I_{2} = -4$$
$$I_{3} = 40 i$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4$$
$$K_{2} = \frac{55}{4}$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 4 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = 2$$
then the canonical form of the equation will be
$$\tilde k^{2} \lambda_{2} + \tilde r^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$2 \tilde k^{2} - 2 \tilde r^{2} - 10 i = 0$$
None
- reduced to canonical form