Algebra Examples

$(1, 2, - 3)$ , $(3, 5, - 3)$ , $(1, - 1, 1)$ , $(- 2, - 2, - 2)$

Step 1

Given points $C = (1, - 1, 1)$ and $D = (- 2, - 2, - 2)$ , find a plane containing points $A = (1, 2, - 3)$ and $B = (3, 5, - 3)$ that is parallel to line $CD$ .

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

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

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

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

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 $CD$ .

$V_{CD} = ⟨ - 3, - 1, - 3 ⟩$

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 $AB$ .

$V_{AB} = ⟨ 2, 3, 0 ⟩$

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

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

Step 7

Calculate the determinant.

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Step 7.1

Choose the row or column with the most $0$ 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.

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

Step 7.1.4

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

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

Step 7.1.5

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

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

Step 7.1.6

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

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

Step 7.1.7

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

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

Step 7.1.8

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

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

Step 7.1.9

Add the terms together.

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

Step 7.2

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

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Step 7.2.1

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

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

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 $3$ by $- 3$ .

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

Step 7.2.2.1.2

Multiply $- - 0$ .

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Step 7.2.2.1.2.1

Multiply $- 1$ by $0$ .

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

Step 7.2.2.1.2.2

Multiply $- 1$ by $0$ .

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

Step 7.2.2.2

Add $- 9$ and $0$ .

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

Step 7.3

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

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Step 7.3.1

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

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

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 $2$ by $- 3$ .

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

Step 7.3.2.1.2

Multiply $- (- 3 \cdot 0)$ .

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Step 7.3.2.1.2.1

Multiply $- 3$ by $0$ .

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

Step 7.3.2.1.2.2

Multiply $- 1$ by $0$ .

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

Step 7.3.2.2

Add $- 6$ and $0$ .

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

Step 7.4

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

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Step 7.4.1

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

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

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 $2$ by $- 1$ .

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

Step 7.4.2.1.2

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

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Step 7.4.2.1.2.1

Multiply $- 3$ by $3$ .

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

Step 7.4.2.1.2.2

Multiply $- 1$ by $- 9$ .

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

Step 7.4.2.2

Add $- 2$ and $9$ .

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

Step 7.5

Simplify each term.

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Step 7.5.1

Move $- 9$ to the left of $i$ .

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

Step 7.5.2

Multiply $- 1$ by $- 6$ .

$- 9 i + 6 j + 7 k$

Step 8

Solve the expression $(- 9) x + (6) y + (7) z$ 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 $- 9$ by $1$ .

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

Step 8.1.2

Multiply $6$ by $2$ .

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

Step 8.1.3

Multiply $7$ by $- 3$ .

$- 9 + 12 - 21$

Step 8.2

Simplify by adding and subtracting.

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Step 8.2.1

Add $- 9$ and $12$ .

$3 - 21$

Step 8.2.2

Subtract $21$ from $3$ .

$- 18$

Step 9

Add the constant to find the equation of the plane to be $(- 9) x + (6) y + (7) z = - 18$ .

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

Step 10

Multiply $7$ by $z$ .

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

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