Answer:
6a. π1 ≡ π3 ║ π2
6b. π2 ≡ π3, intersects π1 in the line ((13+t)/6, (10+16t)/6, t)
7a. k = -4
7b. k = 4
7c. k ≠ ±4
Step-by-step explanation:
You want the relationship between the planes whose equations are provided, and the values of k that make the lines parallel, coincident, intersecting.
6a. Planes
The attachment shows the result from reducing the augmented matrix of equation coefficients to row-echelon form. The second line of the result is 0 0 0 1, indicating the equations describe parallel planes. The third line of the result is 0 0 0 0, indicating two of the planes are coincident.
Putting the plane equations into standard form, we have ...
π1: x -2y +3z = 2
π2: x -2y +3z = -2 . . . . . . parallel to π1
π3: x -2y +3z = 2 . . . . . . . coincident with π1
6b. Planes
Putting the equations for these planes in standard form gives ...
π1: 4x -y +2z = 7
π2: 2x -y +3z = 6 . . . . . . intersects π1
π3: 2x -y +3z = 6 . . . . . . coincident with π2
The second part of the attachment shows the parametric equation of the line of intersection can be written as ...
(x, y, z) = ((13+t)/6, (10+16t)/6, t)
7a. Parallel lines
The determinant of the coefficients of the equations of the lines is ...
[tex]\left|\begin{array}{cc=c}2&k\\k&8\end{array}\right|=16-k^2[/tex]
When this is zero, the lines will be coincident or parallel.
16 -k² = 0
k = ±4 . . . . . . . add k², take the square root
For k = -4, the equations are 2x -4y = 10 and -4x +8y = 20. When written in standard form, these are ...
x -2y = 5
x -2y = -5 . . . . . . parallel lines for k = -4
7b. Coincident lines
For k = 4, the standard form equations are ...
x +2y = 5
x +2y = 5 . . . . . . coincident lines for k = 4
7c. Different lines
For any other values of k, the determinant of the coefficients is not zero, so the lines will intersect in a point.
The lines are different for k ≠ ±4.
