Algebra Examples

$(1, 1, 1)$ , $(1, 2, 3)$ , $(2, 2, 2)$ , $(4, 7, 10)$

Given points $C = (2, 2, 2)$ and $D = (4, 7, 10)$ , find a plane containing points $A = (1, 1, 1)$ and $B = (1, 2, 3)$ that is parallel to line $CD$ .

$A = (1, 1, 1)$

$B = (1, 2, 3)$

$C = (2, 2, 2)$

$D = (4, 7, 10)$

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 $CD$ .

$V_{CD} = 〈 2, 5, 8 〉$

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 $AB$ .

$V_{AB} = 〈 0, 1, 2 〉$

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

$[\begin{array}{c} i & j & k \\ 0 & 1 & 2 \\ 2 & 5 & 8 \end{array}]$

Calculate the determinant of $[\begin{array}{c} i & j & k \\ 0 & 1 & 2 \\ 2 & 5 & 8 \end{array}]$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

$i | \begin{array}{c} 1 & 2 \\ 5 & 8 \end{array} | + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

The determinant of $i | \begin{array}{c} 1 & 2 \\ 5 & 8 \end{array} |$ is $- 2 i$ .

Tap for more steps...

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

$i ((1) (8) - 5 \cdot 2) + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $8$ by $1$ .

$i (8 - 5 \cdot 2) + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Multiply $- 5$ by $2$ .

$i (8 - 10) + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Simplify the expression.

Tap for more steps...

Subtract $10$ from $8$ .

$i \cdot - 2 + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Move $- 2$ to the left of the expression $i \cdot - 2$ .

$- 2 \cdot i + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Multiply $- 2$ by $i$ .

$- 2 i + 0 | \begin{array}{c} j & k \\ 5 & 8 \end{array} | + 2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

The determinant of $2 | \begin{array}{c} j & k \\ 1 & 2 \end{array} |$ is $4 j - 2 k$ .

Tap for more steps...

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

$- 2 i + 0 + 2 ((j) (2) - 1 k)$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $j$ by $2$ .

$- 2 i + 0 + 2 (j (2) - 1 k)$

Multiply $j$ by $2$ .

$- 2 i + 0 + 2 (j \cdot 2 - 1 k)$

Move $2$ to the left of the expression $j \cdot 2$ .

$- 2 i + 0 + 2 (2 \cdot j - 1 k)$

Multiply $2$ by $j$ .

$- 2 i + 0 + 2 (2 j - 1 k)$

Rewrite $- 1 k$ as $- k$ .

$- 2 i + 0 + 2 (2 j - k)$

Simplify by multiplying through.

Tap for more steps...

Apply the distributive property.

$- 2 i + 0 + 2 (2 j) + 2 (- k)$

Multiply.

Tap for more steps...

Multiply $2$ by $2$ .

$- 2 i + 0 + 4 j + 2 (- k)$

Multiply $- 1$ by $2$ .

$- 2 i + 0 + 4 j - 2 k$

Add $- 2 i$ and $0$ .

$- 2 i + 4 j - 2 k$

Solve the expression $(- 2) x + (4) y + (- 2) z$ at point $A$ since it is on the plane. This is used to calculate the constant in the equation for the plane.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $- 2$ by $1$ .

$- 2 + (4) \cdot 1 + (- 2) \cdot 1$

Multiply $4$ by $1$ .

$- 2 + 4 + (- 2) \cdot 1$

Multiply $- 2$ by $1$ .

$- 2 + 4 - 2$

Simplify by adding and subtracting.

Tap for more steps...

Add $- 2$ and $4$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

Add the constant to find the equation of the plane to be $(- 2) x + (4) y + (- 2) z = 0$ .

$(- 2) x + (4) y + (- 2) z = 0$

Simplify each term.

Tap for more steps...

Multiply $- 2$ by $x$ .

$- 2 x + (4) y + (- 2) z = 0$

Multiply $4$ by $y$ .

$- 2 x + 4 y + (- 2) z = 0$

Multiply $- 2$ by $z$ .

$- 2 x + 4 y - 2 z = 0$

Enter YOUR Problem