Examples

(2, 6, - 8)

$(2, 6, - 8)$ ,

$(- 12, - 2, - 1)$ ,

$(- 2, 8, 9)$ ,

$(3, 0, 0)$

Step 1

Given points

$C = (- 2, 8, 9)$ and

$D = (3, 0, 0)$ , find a plane containing points

$A = (2, 6, - 8)$ and

$B = (- 12, - 2, - 1)$ that is parallel to line

$CD$ .

$A = (2, 6, - 8)$

$B = (- 12, - 2, - 1)$

$C = (- 2, 8, 9)$

$D = (3, 0, 0)$

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

$CD$ .

$V_{CD} = ⟨ 5, - 8, - 9 ⟩$

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

$AB$ .

$V_{AB} = ⟨ - 14, - 8, 7 ⟩$

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

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

$[\begin{array}{c} i & j & k \\ - 14 & - 8 & 7 \\ 5 & - 8 & - 9 \end{array}]$

Step 7

Calculate the determinant.

Tap for more steps...

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.

Tap for more steps...

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.

$| \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$

Step 7.1.4

Multiply element

$a_{11}$ by its cofactor.

$i | \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$

Step 7.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} |$

Step 7.1.6

Multiply element

$a_{12}$ by its cofactor.

$- | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j$

Step 7.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} |$

Step 7.1.8

Multiply element

$a_{13}$ by its cofactor.

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.1.9

Add the terms together.

$i | \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} | - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2

Evaluate

$| \begin{array}{c} - 8 & 7 \\ - 8 & - 9 \end{array} |$ .

Tap for more steps...

Step 7.2.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$i (- 8 \cdot - 9 - (- 8 \cdot 7)) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2

Simplify the determinant.

Tap for more steps...

Step 7.2.2.1

Simplify each term.

Tap for more steps...

Step 7.2.2.1.1

Multiply

$- 8$ by

$- 9$ .

$i (72 - (- 8 \cdot 7)) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.1.2

Multiply

$- (- 8 \cdot 7)$ .

Tap for more steps...

Step 7.2.2.1.2.1

Multiply

$- 8$ by

$7$ .

$i (72 - - 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.1.2.2

Multiply

$- 1$ by

$- 56$ .

$i (72 + 56) - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.2.2.2

Add

$72$ and

$56$ .

$i \cdot 128 - | \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} | j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3

Evaluate

$| \begin{array}{c} - 14 & 7 \\ 5 & - 9 \end{array} |$ .

Tap for more steps...

Step 7.3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$i \cdot 128 - (- 14 \cdot - 9 - 5 \cdot 7) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3.2

Simplify the determinant.

Tap for more steps...

Step 7.3.2.1

Simplify each term.

Tap for more steps...

Step 7.3.2.1.1

Multiply

$- 14$ by

$- 9$ .

$i \cdot 128 - (126 - 5 \cdot 7) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3.2.1.2

Multiply

$- 5$ by

$7$ .

$i \cdot 128 - (126 - 35) j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.3.2.2

Subtract

$35$ from

$126$ .

$i \cdot 128 - 1 \cdot 91 j + | \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} | k$

Step 7.4

Evaluate

$| \begin{array}{c} - 14 & - 8 \\ 5 & - 8 \end{array} |$ .

Tap for more steps...

Step 7.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$i \cdot 128 - 1 \cdot 91 j + (- 14 \cdot - 8 - 5 \cdot - 8) k$

Step 7.4.2

Simplify the determinant.

Tap for more steps...

Step 7.4.2.1

Simplify each term.

Tap for more steps...

Step 7.4.2.1.1

Multiply

$- 14$ by

$- 8$ .

$i \cdot 128 - 1 \cdot 91 j + (112 - 5 \cdot - 8) k$

Step 7.4.2.1.2

Multiply

$- 5$ by

$- 8$ .

$i \cdot 128 - 1 \cdot 91 j + (112 + 40) k$

Step 7.4.2.2

Add

$112$ and

$40$ .

$i \cdot 128 - 1 \cdot 91 j + 152 k$

Step 7.5

Simplify each term.

Tap for more steps...

Step 7.5.1

Move

$128$ to the left of

$i$ .

$128 \cdot i - 1 \cdot 91 j + 152 k$

Step 7.5.2

Multiply

$- 1$ by

$91$ .

$128 i - 91 j + 152 k$

Step 8

Solve the expression

$(128) x + (- 91) y + (152) z$ 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...

Step 8.1

Simplify each term.

Tap for more steps...

Step 8.1.1

Multiply

$128$ by

$2$ .

$256 + (- 91) \cdot 6 + (152) \cdot - 8$

Step 8.1.2

Multiply

$- 91$ by

$6$ .

$256 - 546 + (152) \cdot - 8$

Step 8.1.3

Multiply

$152$ by

$- 8$ .

$256 - 546 - 1216$

Step 8.2

Simplify by subtracting numbers.

Tap for more steps...

Step 8.2.1

Subtract

$546$ from

$256$ .

$- 290 - 1216$

Step 8.2.2

Subtract

$1216$ from

$- 290$ .

$- 1506$

Step 9

Add the constant to find the equation of the plane to be

$(128) x + (- 91) y + (152) z = - 1506$ .

$(128) x + (- 91) y + (152) z = - 1506$

Step 10

Multiply

$152$ by

$z$ .

$128 x - 91 y + 152 z = - 1506$

Enter YOUR Problem