Algebra Examples

$(1, 2, 3)$ , $(2, 5, 6)$ , $(2, 9, 7)$ , $(3, 3, 3)$

Given points $C = (2, 9, 7)$ and $D = (3, 3, 3)$ , find a plane containing points $A = (1, 2, 3)$ and $B = (2, 5, 6)$ that is parallel to line $CD$ .

$A = (1, 2, 3)$

$B = (2, 5, 6)$

$C = (2, 9, 7)$

$D = (3, 3, 3)$

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 $D$ .

${Direction Vector}_{CD} = 〈 x (D) - x C, y (D) - y C, z (D) - z C 〉$

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

$V_{CD} = 〈 1, - 6, - 4 〉$

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

${Direction Vector}_{AB} = 〈 x (B) - x A, y (B) - y A, z (B) - z A 〉$

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

$V_{AB} = 〈 1, 3, 3 〉$

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

$[\begin{array}{c} i & j & k \\ 1 & 3 & 3 \\ 1 & - 6 & - 4 \end{array}]$

Calculate the determinant of $[\begin{array}{c} i & j & k \\ 1 & 3 & 3 \\ 1 & - 6 & - 4 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$i | \begin{array}{c} 3 & 3 \\ - 6 & - 4 \end{array} | - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

The determinant of $i | \begin{array}{c} 3 & 3 \\ - 6 & - 4 \end{array} |$ is $6 i$ .

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

$i ((3) (- 4) + 6 \cdot 3) - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 4$ .

$i (- 12 + 6 \cdot 3) - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply $6$ by $3$ .

$i (- 12 + 18) - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Simplify the expression.

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Add $- 12$ and $18$ .

$i \cdot 6 - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Move $6$ to the left of the expression $i \cdot 6$ .

$6 \cdot i - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply $6$ by $i$ .

$6 i - 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} | + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

The determinant of $- 1 | \begin{array}{c} j & k \\ - 6 & - 4 \end{array} |$ is $4 j - 6 k$ .

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

$6 i - 1 ((j) (- 4) + 6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $j$ by $- 4$ .

$6 i - 1 (j (- 4) + 6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply $j$ by $- 4$ .

$6 i - 1 (j \cdot - 4 + 6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Move $- 4$ to the left of the expression $j \cdot - 4$ .

$6 i - 1 (- 4 \cdot j + 6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply $- 4$ by $j$ .

$6 i - 1 (- 4 j + 6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Simplify by multiplying through.

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Apply the distributive property.

$6 i - 1 (- 4 j) - 1 (6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply.

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Multiply $- 4$ by $- 1$ .

$6 i + 4 j - 1 (6 k) + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

Multiply $6$ by $- 1$ .

$6 i + 4 j - 6 k + 1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$

The determinant of $1 | \begin{array}{c} j & k \\ 3 & 3 \end{array} |$ is $3 j - 3 k$ .

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

$6 i + 4 j - 6 k j \cdot 3 - 3 k$

Simplify each term.

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Multiply $j$ by $3$ .

$6 i + 4 j - 6 k + j (3) - 3 k$

Multiply $j$ by $3$ .

$6 i + 4 j - 6 k + j \cdot 3 - 3 k$

Move $3$ to the left of the expression $j \cdot 3$ .

$6 i + 4 j - 6 k + 3 \cdot j - 3 k$

Multiply $3$ by $j$ .

$6 i + 4 j - 6 k + 3 j - 3 k$

Add $4 j$ and $3 j$ .

$6 i + 7 j - 6 k - 3 k$

Subtract $3 k$ from $- 6 k$ .

$6 i + 7 j - 9 k$

Solve the expression $(6) x + (7) y + (- 9) z$ 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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Simplify each term.

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Multiply $6$ by $1$ .

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

Multiply $7$ by $2$ .

$6 + 14 + (- 9) \cdot 3$

Multiply $- 9$ by $3$ .

$6 + 14 - 27$

Simplify by adding and subtracting.

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Add $6$ and $14$ .

$20 - 27$

Subtract $27$ from $20$ .

$- 7$

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

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

Simplify each term.

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Multiply $6$ by $x$ .

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

Multiply $7$ by $y$ .

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

Multiply $- 9$ by $z$ .

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

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