Algebra Examples

$(2, 6, - 8)$ , $(- 12, - 2, - 1)$ , $(- 2, 8, 9)$ , $(3, 0, 0)$

Step 1

Given points $C = (- 2, 8, 9)$ and $D = (3, 0, 0)$ , find a plane containing points $A = (2, 6, - 8)$ and $B = (- 12, - 2, - 1)$ that is parallel to line $CD$ .

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

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

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

$D = (3, 0, 0)$

Step 2

First, calculate the direction vector of the line through points $C$ and $D$ . This can be done by taking the coordinate values of point $C$ and subtracting them from point $D$ .

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

Step 3

Replace the $x$ , $y$ , and $z$ values and then simplify to get the direction vector $V_{CD}$ for line $CD$ .

$V_{CD} = ⟨ 5, - 8, - 9 ⟩$

Step 4

Calculate the direction vector of a line through points $A$ and $B$ using the same method.

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

Step 5

Replace the $x$ , $y$ , and $z$ values and then simplify to get the direction vector $V_{AB}$ for line $AB$ .

$V_{AB} = ⟨ - 14, - 8, 7 ⟩$

Step 6

The solution plane will contain a line that contains points $A$ and $B$ and with the direction vector $V_{A B}$ . For this plane to be parallel to the line $C D$ , find the normal vector of the plane which is also orthogonal to the direction vector of the line $C D$ . Calculate the normal vector by finding the cross product $V_{A B}$ x $V_{C D}$ by finding the determinant of the matrix $[\begin{array}{c} i & j & k \\ x_{B} - x_{A} & y_{B} - y_{A} & z_{B} - z_{A} \\ x_{D} - x_{C} & y_{D} - y_{C} & z_{D} - z_{C} \end{array}]$ .

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

Step 7

Calculate the determinant.

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Step 7.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

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Step 7.1.1

Consider the corresponding sign chart.

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

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 $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 7.1.4

Multiply element $a_{11}$ by its cofactor.

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

Step 7.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

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

Step 7.1.6

Multiply element $a_{12}$ by its cofactor.

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

Step 7.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Step 7.1.8

Multiply element $a_{13}$ by its cofactor.

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

Step 7.1.9

Add the terms together.

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

Step 7.2

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

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Step 7.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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 $- 8$ by $- 9$ .

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

Step 7.2.2.1.2

Multiply $- (- 8 \cdot 7)$ .

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Step 7.2.2.1.2.1

Multiply $- 8$ by $7$ .

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

Step 7.2.2.1.2.2

Multiply $- 1$ by $- 56$ .

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

Step 7.2.2.2

Add $72$ and $56$ .

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

Step 7.3

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

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Step 7.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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 $- 14$ by $- 9$ .

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

Step 7.3.2.1.2

Multiply $- 5$ by $7$ .

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

Step 7.3.2.2

Subtract $35$ from $126$ .

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

Step 7.4

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

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Step 7.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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 $- 14$ by $- 8$ .

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

Step 7.4.2.1.2

Multiply $- 5$ by $- 8$ .

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

Step 7.4.2.2

Add $112$ and $40$ .

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

Step 7.5

Simplify each term.

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Step 7.5.1

Move $128$ to the left of $i$ .

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

Step 7.5.2

Multiply $- 1$ by $91$ .

$128 i - 91 j + 152 k$

Step 8

Solve the expression $(128) x + (- 91) y + (152) z$ at point $A$ 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 $128$ by $2$ .

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

Step 8.1.2

Multiply $- 91$ by $6$ .

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

Step 8.1.3

Multiply $152$ by $- 8$ .

$256 - 546 - 1216$

Step 8.2

Simplify by subtracting numbers.

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Step 8.2.1

Subtract $546$ from $256$ .

$- 290 - 1216$

Step 8.2.2

Subtract $1216$ from $- 290$ .

$- 1506$

Step 9

Add the constant to find the equation of the plane to be $(128) x + (- 91) y + (152) z = - 1506$ .

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

Step 10

Multiply $152$ by $z$ .

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

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