Use normal vectors to determine the intersection, if any, for each of the follow
ID: 3111242 • Letter: U
Question
Use normal vectors to determine the intersection, if any, for each of the following groups of three planes. Give a geometric interpretation in each case and the number of solutions for the corresponding linear system of equations. If the planes intersect in a line, determine a vector equation of the line. If the planes intersect in a point, determine the coordinates of the point.
a. x + 2y + 3z = 4
2x + 4y + 6z = 7
x + 3y + 2z = 3
b. x + 2y + 3z = 4
2x + 4y + 6z = 7
3x + 6y + 9z = 5
c. x + 2y + z = 2
2x + 4y + 2z = 4
3x + 6y + 3z = 6
d. x 2y 2z = 6
2x 5y + 3z = 10
3x 4y + z = 1
e. x y + 3z = 4
x + y + 2z = 2
3x + y + 7z = 9
Explanation / Answer
Solve the system of equations:
x1 + 2x2 + 3x3 = -42x1 + 4x2 + 6x3 = 73x1 + 6x2 + 9x3 = -3
Simplify the system:
x1 + 2x2 + 3x3 = -42x1 + 4x2 + 6x3 = 7x1 + 2x2 + 3x3 = -1
From 1-th equation we find x1 by other variables
x1 = -2x2 - 3x3 - 42x1 + 4x2 + 6x3 = 7x1 + 2x2 + 3x3 = -1
In 2, 3-th equation substitute x1
x1 = -2x2 - 3x3 - 42(-2x2 - 3x3 - 4) + 4x2 + 6x3 = 7(-2x2 - 3x3 - 4) + 2x2 + 3x3 = -1
After simplification we get:
x1 = -2x2 - 3x3 - 40 = 150 = 3
Answer:
The system of equations has no solution because: 0 3.
B)Solve the system of equations:
x1 + 2x2 + 3x3 = -42x1 + 4x2 + 6x3 = 73x1 + 6x2 + 9x3 = 5
From 1-th equation we find x1 by other variables
x1 = -2x2 - 3x3 - 42x1 + 4x2 + 6x3 = 73x1 + 6x2 + 9x3 = 5
In 2, 3-th equation substitute x1
x1 = -2x2 - 3x3 - 42(-2x2 - 3x3 - 4) + 4x2 + 6x3 = 73(-2x2 - 3x3 - 4) + 6x2 + 9x3 = 5
After simplification we get:
x1 = -2x2 - 3x3 - 40 = 150 = 17
Answer:
The system of equations has no solution because: 0 17
C)Solve the system of equations:
x1 + 2x2 + x3 = -22x1 + 4x2 + 2x3 = 43x1 + 6x2 + 3x3 = -6
Simplify the system:
x1 + 2x2 + x3 = -2x1 + 2x2 + x3 = 2x1 + 2x2 + x3 = -2
From 1-th equation we find x1 by other variables
x1 = -2x2 - x3 - 2x1 + 2x2 + x3 = 2x1 + 2x2 + x3 = -2
In 2, 3-th equation substitute x1
x1 = -2x2 - x3 - 2(-2x2 - x3 - 2) + 2x2 + x3 = 2(-2x2 - x3 - 2) + 2x2 + x3 = -2
After simplification we get:
x1 = -2x2 - x3 - 20 = 40 = 0
Answer:
The system of equations has no solution because: 0 4.
D)
Solve the system of equations:
x1 - 2x2 - 2x3 = 62x1 - 5x2 + 3x3 = -103x1 - 4x2 + x3 = -1
From 1-th equation we find x1 by other variables
x1 = 2x2 + 2x3 + 62x1 - 5x2 + 3x3 = -103x1 - 4x2 + x3 = -1
In 2, 3-th equation substitute x1
x1 = 2x2 + 2x3 + 62(2x2 + 2x3 + 6) - 5x2 + 3x3 = -103(2x2 + 2x3 + 6) - 4x2 + x3 = -1
After simplification we get:
x1 = 2x2 + 2x3 + 6-x2 + 7x3 = -222x2 + 7x3 = -19
Divide the 2-th equation by -1
x1 = 2x2 + 2x3 + 6x2 - 7x3 = 222x2 + 7x3 = -19
From 2-th equation we find x2 by other variables
x1 = 2x2 + 2x3 + 6x2 = 7x3 + 222x2 + 7x3 = -19
In 3-th equation substitute x2
x1 = 2x2 + 2x3 + 6x2 = 7x3 + 222(7x3 + 22) + 7x3 = -19
After simplification we get:
x1 = 2x2 + 2x3 + 6x2 = 7x3 + 2221x3 = -63
Divide the 3-th equation by 21
x1 = 2x2 + 2x3 + 6x2 = 7x3 + 22x3 = -3
Now moving from the last equation to first equation we can find the values of other variables
Answer:
x1 = 2x2 = 1x3 = -3.
E )
Solve the system of equations:
x1 - x2 + 3x3 = 4x1 + x2 + 2x3 = 23x1 + x2 + 7x3 = 9
From 1-th equation we find x1 by other variables
x1 = x2 - 3x3 + 4x1 + x2 + 2x3 = 23x1 + x2 + 7x3 = 9
In 2, 3-th equation substitute x1
x1 = x2 - 3x3 + 4(x2 - 3x3 + 4) + x2 + 2x3 = 23(x2 - 3x3 + 4) + x2 + 7x3 = 9
After simplification we get:
x1 = x2 - 3x3 + 42x2 - x3 = -24x2 - 2x3 = -3
Divide the 2-th equation by 2
x1 = x2 - 3x3 + 4x2 - 0.5x3 = -14x2 - 2x3 = -3
From 2-th equation we find x2 by other variables
x1 = x2 - 3x3 + 4x2 = 0.5x3 - 14x2 - 2x3 = -3
In 3-th equation substitute x2
x1 = x2 - 3x3 + 4x2 = 0.5x3 - 14(0.5x3 - 1) - 2x3 = -3
After simplification we get:
x1 = x2 - 3x3 + 4x2 = 0.5x3 - 10 = 1
Answer:
The system of equations has no solution because: 0 1
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