Algebra Examples

$(6, 7, 2)$ , $(- 8, 3, - 1)$ , $(7, 0, - 1)$ , $(0, - 4, 0)$

Step 1

Given points $C = (7, 0, - 1)$ and $D = (0, - 4, 0)$ , find a plane containing points $A = (6, 7, 2)$ and $B = (- 8, 3, - 1)$ that is parallel to line $CD$ .

$A = (6, 7, 2)$

$B = (- 8, 3, - 1)$

$C = (7, 0, - 1)$

$D = (0, - 4, 0)$

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 $CD$ .

$V_{CD} = ⟨ - 7, - 4, 1 ⟩$

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 $AB$ .

$V_{AB} = ⟨ - 14, - 4, - 3 ⟩$

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

$[\begin{array}{c} i & j & k \\ - 14 & - 4 & - 3 \\ - 7 & - 4 & 1 \end{array}]$

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.

$| \begin{array}{c} - 4 & - 3 \\ - 4 & 1 \end{array} |$

Step 7.1.4

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

$i | \begin{array}{c} - 4 & - 3 \\ - 4 & 1 \end{array} |$

Step 7.1.5

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

$| \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} |$

Step 7.1.6

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

$- | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j$

Step 7.1.7

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

$| \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} |$

Step 7.1.8

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

$| \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.1.9

Add the terms together.

$i | \begin{array}{c} - 4 & - 3 \\ - 4 & 1 \end{array} | - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.2

Evaluate $| \begin{array}{c} - 4 & - 3 \\ - 4 & 1 \end{array} |$ .

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Step 7.2.1

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

$i (- 4 \cdot 1 - (- 4 \cdot - 3)) - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

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 $- 4$ by $1$ .

$i (- 4 - (- 4 \cdot - 3)) - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.2.2.1.2

Multiply $- (- 4 \cdot - 3)$ .

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Step 7.2.2.1.2.1

Multiply $- 4$ by $- 3$ .

$i (- 4 - 1 \cdot 12) - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.2.2.1.2.2

Multiply $- 1$ by $12$ .

$i (- 4 - 12) - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.2.2.2

Subtract $12$ from $- 4$ .

$i \cdot - 16 - | \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} | j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.3

Evaluate $| \begin{array}{c} - 14 & - 3 \\ - 7 & 1 \end{array} |$ .

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Step 7.3.1

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

$i \cdot - 16 - (- 14 \cdot 1 - (- 7 \cdot - 3)) j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

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 $1$ .

$i \cdot - 16 - (- 14 - (- 7 \cdot - 3)) j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.3.2.1.2

Multiply $- (- 7 \cdot - 3)$ .

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Step 7.3.2.1.2.1

Multiply $- 7$ by $- 3$ .

$i \cdot - 16 - (- 14 - 1 \cdot 21) j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.3.2.1.2.2

Multiply $- 1$ by $21$ .

$i \cdot - 16 - (- 14 - 21) j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.3.2.2

Subtract $21$ from $- 14$ .

$i \cdot - 16 - - 35 j + | \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} | k$

Step 7.4

Evaluate $| \begin{array}{c} - 14 & - 4 \\ - 7 & - 4 \end{array} |$ .

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Step 7.4.1

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

$i \cdot - 16 - - 35 j + (- 14 \cdot - 4 - (- 7 \cdot - 4)) k$

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 $- 4$ .

$i \cdot - 16 - - 35 j + (56 - (- 7 \cdot - 4)) k$

Step 7.4.2.1.2

Multiply $- (- 7 \cdot - 4)$ .

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Step 7.4.2.1.2.1

Multiply $- 7$ by $- 4$ .

$i \cdot - 16 - - 35 j + (56 - 1 \cdot 28) k$

Step 7.4.2.1.2.2

Multiply $- 1$ by $28$ .

$i \cdot - 16 - - 35 j + (56 - 28) k$

Step 7.4.2.2

Subtract $28$ from $56$ .

$i \cdot - 16 - - 35 j + 28 k$

Step 7.5

Simplify each term.

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Step 7.5.1

Move $- 16$ to the left of $i$ .

$- 16 \cdot i - - 35 j + 28 k$

Step 7.5.2

Multiply $- 1$ by $- 35$ .

$- 16 i + 35 j + 28 k$

Step 8

Solve the expression $(- 16) x + (35) y + (28) z$ 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 $- 16$ by $6$ .

$- 96 + (35) \cdot 7 + (28) \cdot 2$

Step 8.1.2

Multiply $35$ by $7$ .

$- 96 + 245 + (28) \cdot 2$

Step 8.1.3

Multiply $28$ by $2$ .

$- 96 + 245 + 56$

Step 8.2

Simplify by adding numbers.

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Step 8.2.1

Add $- 96$ and $245$ .

$149 + 56$

Step 8.2.2

Add $149$ and $56$ .

$205$

Step 9

Add the constant to find the equation of the plane to be $(- 16) x + (35) y + (28) z = 205$ .

$(- 16) x + (35) y + (28) z = 205$

Step 10

Multiply $28$ by $z$ .

$- 16 x + 35 y + 28 z = 205$

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