given the following plane:
2x-y-6z=5: (x-1, y-1, z-6).(3,1,1)=0; and (x,y,z)=(0,8,9)+s(1,-2,-1)+t(-3,6,3), determine algebraically whether or not this system of equation has a solution.
Let's start by solving the first equation for one of the variables. We choose to solve for y, which gives:
y = 2x - 6z - 5
Now we substitute this expression for y into the second equation:
2x - 6z - 5 - x + 1 - y + 1 - 6z + 3 + t = 0
Simplifying and collecting like terms:
x - 8z + t - 2 = 0
Now we use the third equation to solve for x, y, and z:
x = 0 + s(1) + t(-3)
y = 8 + s(-2) + t(6)
z = 9 + s(-1) + t(3)
Substituting these expressions into the previous equation, we get:
(0 + s - 3t) - 8(9 - s + 3t) + t - 2 = 0
Simplifying and collecting like terms:
-23s - 59t - 70 = 0
Now we have two linear equations in two variables:
x - 8z + t - 2 = 0
-23s - 59t - 70 = 0
Solving for t in terms of s from the second equation:
t = (-23/59)s + 70/59
Substituting this expression into the first equation:
x - 8z + (-23/59)s + 70/59 - 2 = 0
Simplifying and collecting like terms:
x - 8z + (-23/59)s + 8/59 = 0
Now we substitute the expressions for x and z in terms of s and t:
(0 + s - 3t) - 8(9 - s + 3t) + (-23/59)s + 8/59 = 0
Simplifying and collecting like terms:
-169s + 501t - 450 = 0
Substituting the expression for t in terms of s:
-169s + 501((-23/59)s + 70/59) - 450 = 0
Simplifying and solving for s:
s = 10
Now we can substitute this value of s into the expressions for x, y, and z:
x = 0 + 10 - 3t = 10 - 3t
y = 8 - 2(10) + 6t = -12 + 6t
z = 9 - 10 + 3t = -1 + 3t
Substituting these expressions into the equation of the plane:
2(10 - 3t) - (-12 + 6t) - 6(-1 + 3t) = 5
Simplifying and solving for t:
23t = 14
t = 14/23
Therefore, the system of equations has a unique solution, given by:
x = 10 - 3(14/23) = 20/23
y = -12 + 6(14/23) = -24/23
z = -1 + 3(14/23) = 31/23