Algebra Examples

(1, 2, - 3)

$(1, 2, - 3)$ ,

(3, 5, - 3)

$(3, 5, - 3)$ ,

(1, - 1, 1)

$(1, - 1, 1)$ ,

(- 2, - 2, - 2)

$(- 2, - 2, - 2)$

Step 1

Given points

C = (1, - 1, 1)

$C = (1, - 1, 1)$ and

D = (- 2, - 2, - 2)

$D = (- 2, - 2, - 2)$ , find a plane containing points

A = (1, 2, - 3)

$A = (1, 2, - 3)$ and

B = (3, 5, - 3)

$B = (3, 5, - 3)$ that is parallel to line

CD

$CD$ .

A = (1, 2, - 3)

$A = (1, 2, - 3)$

B = (3, 5, - 3)

$B = (3, 5, - 3)$

C = (1, - 1, 1)

$C = (1, - 1, 1)$

D = (- 2, - 2, - 2)

$D = (- 2, - 2, - 2)$

Step 2

First, calculate the direction vector of the line through points

C

$C$ and

D

$D$ . This can be done by taking the coordinate values of point

C

$C$ and subtracting them from point

D

$D$ .

V_{C D} = < x_{D} - x_{C}, y_{D} - y_{C}, z_{D} - z_{C} >

$V_{C D} = < x_{D} - x_{C}, y_{D} - y_{C}, z_{D} - z_{C} >$

Step 3

Replace the

x

$x$ ,

y

$y$ , and

z

$z$ values and then simplify to get the direction vector

V_{CD}

$V_{CD}$ for line

CD

$CD$ .

V_{CD} = ⟨ - 3, - 1, - 3 ⟩

$V_{CD} = ⟨ - 3, - 1, - 3 ⟩$

Step 4

Calculate the direction vector of a line through points

A

$A$ and

B

$B$ using the same method.

V_{A B} = < x_{B} - x_{A}, y_{B} - y_{A}, z_{B} - z_{A} >

$V_{A B} = < x_{B} - x_{A}, y_{B} - y_{A}, z_{B} - z_{A} >$

Step 5

Replace the

x

$x$ ,

y

$y$ , and

z

$z$ values and then simplify to get the direction vector

V_{AB}

$V_{AB}$ for line

AB

$AB$ .

V_{AB} = ⟨ 2, 3, 0 ⟩

$V_{AB} = ⟨ 2, 3, 0 ⟩$

Step 6

The solution plane will contain a line that contains points

A

$A$ and

B

$B$ and with the direction vector

V_{A B}

$V_{A B}$ . For this plane to be parallel to the line

C D

$C D$ , find the normal vector of the plane which is also orthogonal to the direction vector of the line

C D

$C D$ . Calculate the normal vector by finding the cross product

V_{A B}

$V_{A B}$ x

V_{C D}

$V_{C D}$ by finding the determinant of the matrix

[\begin{array}{c} i & j & k \\ x_{B} - x_{A} & y_{B} - y_{A} & z_{B} - z_{A} \\ x_{D} - x_{C} & y_{D} - y_{C} & z_{D} - z_{C} \end{array}]

$[\begin{array}{c} i & j & k \\ x_{B} - x_{A} & y_{B} - y_{A} & z_{B} - z_{A} \\ x_{D} - x_{C} & y_{D} - y_{C} & z_{D} - z_{C} \end{array}]$ .

[\begin{array}{c} i & j & k \\ 2 & 3 & 0 \\ - 3 & - 1 & - 3 \end{array}]

$[\begin{array}{c} i & j & k \\ 2 & 3 & 0 \\ - 3 & - 1 & - 3 \end{array}]$

Step 7

Calculate the determinant.

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Step 7.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

1

$1$ by its cofactor and add.

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Step 7.1.1

Consider the corresponding sign chart.

| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 7.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Step 7.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

| \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |

$| \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |$

Step 7.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |

$i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |$

Step 7.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

| \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} |

$| \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} |$

Step 7.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j

$- | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j$

Step 7.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} |

$| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} |$

Step 7.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.1.9

Add the terms together.

i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} | - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} | - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} | - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i | \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} | - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.2

Evaluate

| \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |

$| \begin{array}{c} 3 & 0 \\ - 1 & - 3 \end{array} |$ .

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Step 7.2.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

i (3 \cdot - 3 - - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (3 \cdot - 3 - - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.2.2

Simplify the determinant.

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Step 7.2.2.1

Simplify each term.

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Step 7.2.2.1.1

Multiply

3

$3$ by

- 3

$- 3$ .

i (- 9 - - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (- 9 - - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.2.2.1.2

Multiply

- - 0

$- - 0$ .

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Step 7.2.2.1.2.1

Multiply

- 1

$- 1$ by

0

$0$ .

i (- 9 - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (- 9 - 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.2.2.1.2.2

Multiply

- 1

$- 1$ by

0

$0$ .

i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i (- 9 + 0) - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.2.2.2

Add

- 9

$- 9$ and

0

$0$ .

i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - | \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} | j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.3

Evaluate

| \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} |

$| \begin{array}{c} 2 & 0 \\ - 3 & - 3 \end{array} |$ .

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Step 7.3.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

i \cdot - 9 - (2 \cdot - 3 - (- 3 \cdot 0)) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (2 \cdot - 3 - (- 3 \cdot 0)) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.3.2

Simplify the determinant.

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Step 7.3.2.1

Simplify each term.

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Step 7.3.2.1.1

Multiply

2

$2$ by

- 3

$- 3$ .

i \cdot - 9 - (- 6 - (- 3 \cdot 0)) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (- 6 - (- 3 \cdot 0)) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.3.2.1.2

Multiply

- (- 3 \cdot 0)

$- (- 3 \cdot 0)$ .

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Step 7.3.2.1.2.1

Multiply

- 3

$- 3$ by

0

$0$ .

i \cdot - 9 - (- 6 - 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (- 6 - 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.3.2.1.2.2

Multiply

- 1

$- 1$ by

0

$0$ .

i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - (- 6 + 0) j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.3.2.2

Add

- 6

$- 6$ and

0

$0$ .

i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k

$i \cdot - 9 - - 6 j + | \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} | k$

Step 7.4

Evaluate

| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} |

$| \begin{array}{c} 2 & 3 \\ - 3 & - 1 \end{array} |$ .

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Step 7.4.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

i \cdot - 9 - - 6 j + (2 \cdot - 1 - (- 3 \cdot 3)) k

$i \cdot - 9 - - 6 j + (2 \cdot - 1 - (- 3 \cdot 3)) k$

Step 7.4.2

Simplify the determinant.

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Step 7.4.2.1

Simplify each term.

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Step 7.4.2.1.1

Multiply

2

$2$ by

- 1

$- 1$ .

i \cdot - 9 - - 6 j + (- 2 - (- 3 \cdot 3)) k

$i \cdot - 9 - - 6 j + (- 2 - (- 3 \cdot 3)) k$

Step 7.4.2.1.2

Multiply

- (- 3 \cdot 3)

$- (- 3 \cdot 3)$ .

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Step 7.4.2.1.2.1

Multiply

- 3

$- 3$ by

3

$3$ .

i \cdot - 9 - - 6 j + (- 2 - - 9) k

$i \cdot - 9 - - 6 j + (- 2 - - 9) k$

Step 7.4.2.1.2.2

Multiply

- 1

$- 1$ by

- 9

$- 9$ .

i \cdot - 9 - - 6 j + (- 2 + 9) k

$i \cdot - 9 - - 6 j + (- 2 + 9) k$

i \cdot - 9 - - 6 j + (- 2 + 9) k

$i \cdot - 9 - - 6 j + (- 2 + 9) k$

i \cdot - 9 - - 6 j + (- 2 + 9) k

$i \cdot - 9 - - 6 j + (- 2 + 9) k$

Step 7.4.2.2

Add

- 2

$- 2$ and

9

$9$ .

i \cdot - 9 - - 6 j + 7 k

$i \cdot - 9 - - 6 j + 7 k$

i \cdot - 9 - - 6 j + 7 k

$i \cdot - 9 - - 6 j + 7 k$

i \cdot - 9 - - 6 j + 7 k

$i \cdot - 9 - - 6 j + 7 k$

Step 7.5

Simplify each term.

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Step 7.5.1

Move

- 9

$- 9$ to the left of

i

$i$ .

- 9 \cdot i - - 6 j + 7 k

$- 9 \cdot i - - 6 j + 7 k$

Step 7.5.2

Multiply

- 1

$- 1$ by

- 6

$- 6$ .

- 9 i + 6 j + 7 k

$- 9 i + 6 j + 7 k$

- 9 i + 6 j + 7 k

$- 9 i + 6 j + 7 k$

- 9 i + 6 j + 7 k

$- 9 i + 6 j + 7 k$

Step 8

Solve the expression

(- 9) x + (6) y + (7) z

$(- 9) x + (6) y + (7) z$ at point

A

$A$ since it is on the plane. This is used to calculate the constant in the equation for the plane.

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Step 8.1

Simplify each term.

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Step 8.1.1

Multiply

- 9

$- 9$ by

1

$1$ .

- 9 + (6) \cdot 2 + (7) \cdot - 3

$- 9 + (6) \cdot 2 + (7) \cdot - 3$

Step 8.1.2

Multiply

6

$6$ by

2

$2$ .

- 9 + 12 + (7) \cdot - 3

$- 9 + 12 + (7) \cdot - 3$

Step 8.1.3

Multiply

7

$7$ by

- 3

$- 3$ .

- 9 + 12 - 21

$- 9 + 12 - 21$

- 9 + 12 - 21

$- 9 + 12 - 21$

Step 8.2

Simplify by adding and subtracting.

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Step 8.2.1

Add

- 9

$- 9$ and

12

$12$ .

3 - 21

$3 - 21$

Step 8.2.2

Subtract

21

$21$ from

3

$3$ .

- 18

$- 18$

- 18

$- 18$

- 18

$- 18$

Step 9

Add the constant to find the equation of the plane to be

(- 9) x + (6) y + (7) z = - 18

$(- 9) x + (6) y + (7) z = - 18$ .

(- 9) x + (6) y + (7) z = - 18

$(- 9) x + (6) y + (7) z = - 18$

Step 10

Multiply

7

$7$ by

z

$z$ .

- 9 x + 6 y + 7 z = - 18

$- 9 x + 6 y + 7 z = - 18$

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