Question

In: Advanced Math

Use normal vectors to determine the intersection, if any, for each of the following groups of...

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

b x + 2y + 3z = −4

2x + 4y + 6z = 7

2x + 4y + 6z = 7

x + 3y + 2z = −3

3x + 6y + 9z = 5

c x + 2y + z = −2

d x − 2y − 2z = 6 2x + 4y + 2z = 4 2x − 5y + 3z = −10 3x + 6y + 3z = −6 3x − 4y + z = −1

e. x − y + 3z = 4 x + y + 2z = 2 3x + y + 7z = 9

Solutions

Expert Solution

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

Colby Messinger answered 1 year ago
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