# Thread: Equations of Planes 5

1. ## Equations of Planes 5

Find the scalar equation of the plane containing the parallel lines:
r=(2,2,-1)+s(3,-3,1) and r=(1,2,-3)+t(3,-3,1)

Can someone check my work for accurateness? I don't have the answer to this question and would like to know if it is correct. Thanks for your help

P(2,2,-1)
Q(1,2,-3)
d3=PQ=(-1,0,-2)
n=d1 x d3
n=(6,5,-3)
6x+5y-3z+D=0
6(1)+5(2)-3(-3)+D=0
D=-25

6x+5y-3z-25=0

2. Hello, skeske1234!

Find the scalar equation of the plane containing the parallel lines:
. . $\displaystyle r\:=\:(2,2,-1)+s(3,-3,1)\:\text{ and }\:r\:=\:(1,2,-3)+t(3,-3,1)$

Can someone check my work for accurateness?
I don't have the answer to this question and would like to know if it is correct.

$\displaystyle P(2,2,-1),\;Q(1,2,-3)$

$\displaystyle d_3\:=\:PQ\:=\:(-1,0,-2)$

$\displaystyle n\:=\:d_1 \times d_3 \:=\:(6,5,-3)$

$\displaystyle 6x+5y-3z+D\:=\:0$
$\displaystyle 6(1)+5(2)-3(-3)+D\:=\:0 \quad\Rightarrow\quad D\:=\:-25$

Final answer: .$\displaystyle 6x+5y-3z-25\:=\:0$
Correct!

Are the two lines contained in your plane?

The lines are: .$\displaystyle L_1\;\begin{Bmatrix}x &=& 2 + 3s \\ y &=& 2 - 3s \\ z &=& \text{-}1 + s \end{Bmatrix} \qquad L_2\;\begin{Bmatrix}x &=& 1 + 3t \\ y &=& 2 - 3t \\ z &=& \text{-}3 + t \end{Bmatrix}$

Is $\displaystyle L_1$ in your plane: .$\displaystyle 6x + 5y - 3z - 25 \:=\:0$ ?

. . $\displaystyle \begin{array}{cccc}6(2+3s) + 5(2-3s) - 3(\text{-}1 + s) - 25 \:=\:0 & & \text{Is this true?} \\ 12 + 18s + 10 - 15s + 3 - 3s - 25 \:=\:0 \\ 0 \:=\:0 & & \text{Yes!}\end{array}$

Is $\displaystyle L_2$ in your plane: .$\displaystyle 6x + 5y - 3z - 25 \:=\:0$ ?

. . $\displaystyle \begin{array}{cccc}6(1+3t) + 5(2-3t) - 3(\text{-}3 + t) - 25 \:=\:0 & & \text{Is this true?} \\ 6 + 18t + 10 - 15t + 9 - 3t - 25 \:=\:0 \\ 0 \:=\:0 & & \text{Yes!}\end{array}$

Both lines are contained in your plane.