Algebra Examples

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

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 $CD$ .

$A = (6, 7, 2)$

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

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

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

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 to get $〈 - (0 + 7), - (- 4 + 0), - (0 - 1) 〉$ and then simplify to get the direction vector $V_{CD}$ for line $CD$ .

$V_{CD} = 〈 - 7, - 4, 1 〉$

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 to get $〈 - (- 8 + 6), - (3 + 7), - (- 1 + 2) 〉$ and then simplify to get the direction vector $V_{AB}$ for line $AB$ .

$V_{AB} = 〈 - 14, - 4, - 3 〉$

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

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

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

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Set up the determinant by breaking it into smaller components.

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

The determinant of $i | \begin{array}{c} - 4 & - 3 \\ - 4 & 1 \end{array} |$ is $- 16 i$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $- 4$ by $1$ to get $- 4$ .

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

Multiply $4$ by $- 3$ to get $- 12$ .

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

Simplify the expression.

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Subtract $12$ from $- 4$ to get $- 16$ .

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

Move $- 16$ to the left of the expression $i \cdot - 16$ .

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

Multiply $- 16$ by $i$ to get $- 16 i$ .

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

The determinant of $14 | \begin{array}{c} j & k \\ - 4 & 1 \end{array} |$ is $14 j + 56 k$ .

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

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

Simplify the determinant.

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Multiply $j$ by $1$ to get $j$ .

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

Apply the distributive property.

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

Multiply $4$ by $14$ to get $56$ .

$- 16 i + 14 j + 56 k - 7 | \begin{array}{c} j & k \\ - 4 & - 3 \end{array} |$

The determinant of $- 7 | \begin{array}{c} j & k \\ - 4 & - 3 \end{array} |$ is $21 j - 28 k$ .

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

$- 16 i + 14 j + 56 k - 7 ((j) (- 3) + 4 k)$

Simplify the determinant.

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Simplify each term.

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Multiply $j$ by $- 3$ to get $j (- 3)$ .

$- 16 i + 14 j + 56 k - 7 (j (- 3) + 4 k)$

Multiply $j$ by $- 3$ to get $j \cdot - 3$ .

$- 16 i + 14 j + 56 k - 7 (j \cdot - 3 + 4 k)$

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

$- 16 i + 14 j + 56 k - 7 (- 3 \cdot j + 4 k)$

Multiply $- 3$ by $j$ to get $- 3 j$ .

$- 16 i + 14 j + 56 k - 7 (- 3 j + 4 k)$

Simplify by multiplying through.

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Apply the distributive property.

$- 16 i + 14 j + 56 k - 7 (- 3 j) - 7 (4 k)$

Multiply.

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Multiply $- 3$ by $- 7$ to get $21$ .

$- 16 i + 14 j + 56 k + 21 j - 7 (4 k)$

Multiply $4$ by $- 7$ to get $- 28$ .

$- 16 i + 14 j + 56 k + 21 j - 28 k$

Add $14 j$ and $21 j$ to get $35 j$ .

$- 16 i + 35 j + 56 k - 28 k$

Subtract $28 k$ from $56 k$ to get $28 k$ .

$- 16 i + 35 j + 28 k$

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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Simplify each term.

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Multiply $- 16$ by $6$ to get $- 96$ .

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

Multiply $35$ by $7$ to get $245$ .

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

Multiply $28$ by $2$ to get $56$ .

$- 96 + 245 + 56$

Simplify by adding numbers.

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Add $- 96$ and $245$ to get $149$ .

$149 + 56$

Add $149$ and $56$ to get $205$ .

$205$

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

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

Simplify each term.

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Multiply $- 16$ by $x$ to get $- 16 x$ .

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

Multiply $35$ by $y$ to get $35 y$ .

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

Multiply $28$ by $z$ to get $28 z$ .

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

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