Solve the problem.
-Find the volume of the parallelepiped enclosed by the planes 9x + y - z = 1, 9x + y - z = 3, -x + 9y + z = -2, -x + 9y + z = 5, x - y + 3z = -1, x - y + 3z = 6.
A) $\frac { 7 } { 66 }$
B) $\frac { 7 } { 132 }$
C) $\frac { 49 } { 132 }$
D) $\frac { 7 } { 88 }$

Question

Solve the problem.
-Find the volume of the parallelepiped enclosed by the planes 9x + y - z = 1, 9x + y - z = 3, -x + 9y + z = -2, -x + 9y + z = 5, x - y + 3z = -1, x - y + 3z = 6. 
A) $\frac { 7 } { 66 }$
B) $\frac { 7 } { 132 }$
C) $\frac { 49 } { 132 }$
D) $\frac { 7 } { 88 }$

Quizplus · Accepted Answer

First, we need to find the intersection points of the planes. We can do this by setting up a system of equations and solving:

9x + y - z = 1
9x + y - z = 3

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 2

This means that the two planes are parallel and do not intersect.

-x + 9y + z = -2
-x + 9y + z = 5

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 7

This means that the two planes are parallel and do not intersect.

x - y + 3z = -1
x - y + 3z = 6

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 7

This means that the two planes are parallel and do not intersect.

So we only have three planes that intersect. We can find the intersection points by setting up the following system of equations and solving:

9x + y - z = 1
-x + 9y + z = -2
x - y + 3z = -1

We can use substitution to eliminate variables and solve for one variable at a time:

From the first equation, we get:

z = 9x + y - 1

Substituting this into the second equation, we get:

-x + 9y + 9x + y - 1 = -2

Simplifying and solving for y, we get:

y = -\frac {1}{8} x - \frac {1}{8}

Substituting these into the third equation, we get:

x - (-\frac {1}{8} x - \frac {1}{8}) + 3(9x + y - 1) = -1

Simplifying and solving for x, we get:

x = \frac {1}{13}

Substituting these values back into the equations, we can find the corresponding values of y and z:

y = -\frac {1}{13}
z = \frac {116}{169}

Now we can find the vectors that correspond to two edges of the parallelepiped by subtracting the coordinates of two of the intersection points:

\vec{v_1} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {1}{13}, -\frac {1}{8}, \frac {41}{104} angle = \langle 0, \frac {3}{52}, \frac {73}{338} angle

\vec{v_2} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {3}{40}, -\frac {1}{13}, \frac {41}{104} angle = \langle -\frac {11}{520}, \frac {9}{520}, \frac {27}{1690} angle

The volume of the parallelepiped is given by the scalar triple product of the three vectors that correspond to the edges of the parallelepiped:

V = \vec{v_1} \cdot (\vec{v_2} 	imes \vec{v_3})

where \vec{v_3} is the vector that corresponds to the third edge of the parallelepiped. We can find \vec{v_3} in a similar way to \vec{v_2}:

\vec{v_3} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {7}{40}, -\frac {1}{13}, \frac {15}{104} angle = \langle -\frac {5}{52}, \frac {3}{52}, \frac {23}{676} angle

The scalar triple product is:

\vec{v_1} \cdot (\vec{v_2} 	imes \vec{v_3}) = \begin{vmatrix} 0 & \frac {3}{52} & \frac {73}{338} \ -\frac {11}{520} & \frac {9}{520} & \frac {27}{1690} \ -\frac {5}{52} & \frac {3}{52} & \frac {23}{676} \end{vmatrix} = -\frac {49}{1320}

Therefore, the volume of the parallelepiped is:

V = \left| -\frac {49}{1320} ight| = \boxed{\frac {49}{132}}

Solve the Problem 766\frac { 7 } { 66 }667​ B) 7132\frac { 7 } { 132 }1327​

Solve the Problem $\frac { 7 } { 66 }$ B) $\frac { 7 } { 132 }$