The equation of the plane that goes through these points is:
6x + 2y + z = 10.
<h3>How to find the equation of a plane given three points?</h3>
The equation of the plane is found replacing the points into the following equation:
ax + by + c = z.
For point A, we have that:
3b + c = 4.
For point B, we have that:
a + 2b + c = 0.
For point C, we have that:
-a + 6b + c = 4.
Hence the system is:
From the first equation, we have that:
c = 4 - 3b.
Replacing in the second, we have that:
a + 2b + 4 - 3b = 0
a - b = -4.
Replacing in the third, we have that:
-a + 6b + 4 - 3b = 4.
-a + 3b = 0.
a = 3b.
We have that a - b = -4, hence:
3b - b = -4
2b = -4
b = -2.
a = 3b, hence a = -6.
c = 4 - 3b -> c = 10.
Hence the equation is:
ax + by + c = z.
z = -6x - 2y + 10
6x + 2y + z = 10.
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Answer:
Step-by-step explanation:
Let 
. We have that 
if and only if we can find scalars 
such that 
. This can be translated to the following equations:
1. 
2.
3. 
Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for 
and check if the third equationd is fulfilled.
Case (2,6,6)
Using equations 1 and 2 we get
whose unique solutions are 
, but note that for this values, the third equation doesn't hold (3+2 = 5 
6). So this vector is not in the generated space of u and v.
Case (-9,-2,5)
Using equations 1 and 2 we get
whose unique solutions are 
. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.
8 is the x intercept and -2 is the y
2x+y-3=0,x-2y=1 okay that's the answer good luck
