Examples

(2, 6, - 8)

$(2, 6, - 8)$ ,

(- 12, - 2, - 1)

$(- 12, - 2, - 1)$ ,

(- 2, 8, 9)

$(- 2, 8, 9)$ ,

(3, 0, 0)

$(3, 0, 0)$

Step 1

Given points

C = (- 2, 8, 9)

$C = (- 2, 8, 9)$ and

D = (3, 0, 0)

$D = (3, 0, 0)$ , find a plane containing points

A = (2, 6, - 8)

$A = (2, 6, - 8)$ and

B = (- 12, - 2, - 1)

$B = (- 12, - 2, - 1)$ that is parallel to line

CD

$CD$ .

A = (2, 6, - 8)

$A = (2, 6, - 8)$

B = (- 12, - 2, - 1)

$B = (- 12, - 2, - 1)$

C = (- 2, 8, 9)

$C = (- 2, 8, 9)$

D = (3, 0, 0)

$D = (3, 0, 0)$

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} = ⟨ 5, - 8, - 9 ⟩

$V_{CD} = ⟨ 5, - 8, - 9 ⟩$

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} = ⟨ - 14, - 8, 7 ⟩

$V_{AB} = ⟨ - 14, - 8, 7 ⟩$

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{matrix} 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{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{matrix} i & j & k - 14 & - 8 & 7 5 & - 8 & - 9 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} i & j & k \\ - 14 & - 8 & 7 \\ 5 & - 8 & - 9 \end{array}]$

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{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

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

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{matrix} - 8 & 7 - 8 & - 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$

Step 7.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

i ∣ ∣ ∣ \begin{matrix} - 8 & 7 - 8 & - 9 \end{matrix} ∣ ∣ ∣

$i | \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$

Step 7.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} |$

Step 7.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j

$- | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j$

Step 7.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} |$

Step 7.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.1.9

Add the terms together.

i ∣ ∣ ∣ \begin{matrix} - 8 & 7 - 8 & - 9 \end{matrix} ∣ ∣ ∣ - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i | \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} | - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i ∣ ∣ ∣ \begin{matrix} - 8 & 7 - 8 & - 9 \end{matrix} ∣ ∣ ∣ - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i | \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} | - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2

Evaluate

∣ ∣ ∣ \begin{matrix} - 8 & 7 - 8 & - 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$ .

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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{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

i (- 8 \cdot - 9 - (- 8 \cdot 7)) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (- 8 \cdot - 9 - (- 8 \cdot 7)) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

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

- 8

$- 8$ by

- 9

$- 9$ .

i (72 - (- 8 \cdot 7)) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (72 - (- 8 \cdot 7)) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.1.2

Multiply

- (- 8 \cdot 7)

$- (- 8 \cdot 7)$ .

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Step 7.2.2.1.2.1

Multiply

- 8

$- 8$ by

7

$7$ .

i (72 - - 56) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (72 - - 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.1.2.2

Multiply

- 1

$- 1$ by

- 56

$- 56$ .

i (72 + 56) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (72 + 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i (72 + 56) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (72 + 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i (72 + 56) - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i (72 + 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.2

Add

72

$72$ and

56

$56$ .

i \cdot 128 - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i \cdot 128 - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i \cdot 128 - ∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣ j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3

Evaluate

∣ ∣ ∣ \begin{matrix} - 14 & 7 5 & - 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} |$ .

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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{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

i \cdot 128 - (- 14 \cdot - 9 - 5 \cdot 7) j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - (- 14 \cdot - 9 - 5 \cdot 7) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

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

- 14

$- 14$ by

- 9

$- 9$ .

i \cdot 128 - (126 - 5 \cdot 7) j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - (126 - 5 \cdot 7) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3.2.1.2

Multiply

- 5

$- 5$ by

7

$7$ .

i \cdot 128 - (126 - 35) j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - (126 - 35) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i \cdot 128 - (126 - 35) j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - (126 - 35) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3.2.2

Subtract

35

$35$ from

126

$126$ .

i \cdot 128 - 1 \cdot 91 j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - 1 \cdot 91 j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i \cdot 128 - 1 \cdot 91 j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - 1 \cdot 91 j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

i \cdot 128 - 1 \cdot 91 j + ∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣ k

$i \cdot 128 - 1 \cdot 91 j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.4

Evaluate

∣ ∣ ∣ \begin{matrix} - 14 & - 8 5 & - 8 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} |$ .

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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{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

i \cdot 128 - 1 \cdot 91 j + (- 14 \cdot - 8 - 5 \cdot - 8) k

$i \cdot 128 - 1 \cdot 91 j + (- 14 \cdot - 8 - 5 \cdot - 8) k$

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

- 14

$- 14$ by

- 8

$- 8$ .

i \cdot 128 - 1 \cdot 91 j + (112 - 5 \cdot - 8) k

$i \cdot 128 - 1 \cdot 91 j + (112 - 5 \cdot - 8) k$

Step 7.4.2.1.2

Multiply

- 5

$- 5$ by

- 8

$- 8$ .

i \cdot 128 - 1 \cdot 91 j + (112 + 40) k

$i \cdot 128 - 1 \cdot 91 j + (112 + 40) k$

i \cdot 128 - 1 \cdot 91 j + (112 + 40) k

$i \cdot 128 - 1 \cdot 91 j + (112 + 40) k$

Step 7.4.2.2

Add

112

$112$ and

40

$40$ .

i \cdot 128 - 1 \cdot 91 j + 152 k

$i \cdot 128 - 1 \cdot 91 j + 152 k$

i \cdot 128 - 1 \cdot 91 j + 152 k

$i \cdot 128 - 1 \cdot 91 j + 152 k$

i \cdot 128 - 1 \cdot 91 j + 152 k

$i \cdot 128 - 1 \cdot 91 j + 152 k$

Step 7.5

Simplify each term.

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Step 7.5.1

Move

128

$128$ to the left of

i

$i$ .

128 \cdot i - 1 \cdot 91 j + 152 k

$128 \cdot i - 1 \cdot 91 j + 152 k$

Step 7.5.2

Multiply

- 1

$- 1$ by

91

$91$ .

128 i - 91 j + 152 k

$128 i - 91 j + 152 k$

128 i - 91 j + 152 k

$128 i - 91 j + 152 k$

128 i - 91 j + 152 k

$128 i - 91 j + 152 k$

Step 8

Solve the expression

(128) x + (- 91) y + (152) z

$(128) x + (- 91) y + (152) z$ 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

128

$128$ by

2

$2$ .

256 + (- 91) \cdot 6 + (152) \cdot - 8

$256 + (- 91) \cdot 6 + (152) \cdot - 8$

Step 8.1.2

Multiply

- 91

$- 91$ by

6

$6$ .

256 - 546 + (152) \cdot - 8

$256 - 546 + (152) \cdot - 8$

Step 8.1.3

Multiply

152

$152$ by

- 8

$- 8$ .

256 - 546 - 1216

$256 - 546 - 1216$

256 - 546 - 1216

$256 - 546 - 1216$

Step 8.2

Simplify by subtracting numbers.

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Step 8.2.1

Subtract

546

$546$ from

256

$256$ .

- 290 - 1216

$- 290 - 1216$

Step 8.2.2

Subtract

1216

$1216$ from

- 290

$- 290$ .

- 1506

$- 1506$

- 1506

$- 1506$

- 1506

$- 1506$

Step 9

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

(128) x + (- 91) y + (152) z = - 1506

$(128) x + (- 91) y + (152) z = - 1506$ .

(128) x + (- 91) y + (152) z = - 1506

$(128) x + (- 91) y + (152) z = - 1506$

Step 10

Multiply

152

$152$ by

z

$z$ .

128 x - 91 y + 152 z = - 1506

$128 x - 91 y + 152 z = - 1506$

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