Equation of plane through two points and parallel to line

Quote:

Find the equation of the plane passing through the points (3, 2, -1) and (1, -1, 2) that is parallel to the line r(t) = <1, -1, 0> + t<3, 2, -1>

Let P = (3, 2, -1) , Q = (1, -1, 2) and R = <3, 2, -1> = direction vector from r(t)

a = PQ = <1, -1, 2> - <3, 2, -1> = <-2, 3, -3>

b = QR = <3, 2, -1> - <1, -1, 2> = <2, 3, -3>

N = a x b = <0, -12, -12>

Now I'm going to choose a point on the plane, P = (3, 2, -1)

d = n dot OP = <0, -12, -12> dot <3, 2, -1> = -12(2) + 12 = -12

So it follows that the plane has equation -12y + -12z = -12 or y + z = 1 **So it appears that using the direction vector, as I did, from r(t) and generating the plane using those three points is completely wrong. Any guidance?**

Re: Equation of plane through two points and parallel to line

Hi, could you please explain this question, i am completely stuck on it. Thanks!!

Re: Equation of plane through two points and parallel to line

Hi, would you be able to re-explain this question from the beginning, i have to understand it for school tomorrow and Im so confused.

Re: Equation of plane through two points and parallel to line

A plane with equation $\displaystyle \displaystyle \begin{align*} a\,x + b\,y + c\,z = d \end{align*}$ has the same coefficients as the normal vector $\displaystyle \displaystyle \begin{align*} \mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \end{align*}$. To get a normal vector to the plane, you take the cross product of two vectors in the plane. You already have one vector $\displaystyle \displaystyle \begin{align*} \mathbf{r} \end{align*}$. You can use the two points you are given to get another vector in the plane.

Go from here.

Re: Equation of plane through two points and parallel to line

So, I take PQ X R as my coefficients of the plane?

(-2,-3,3)x(3,2,-1)=(-3,7,5)

-3x+7y+5z=0 is parallel to the line and includes p and q?

Re: Equation of plane through two points and parallel to line

and lol at the song lyrics

Re: Equation of plane through two points and parallel to line

P, Q, and X are in a plane perpendicular to r(t) in dir **d**

(PQ)X(PX) is perpendicular to plane and has to be perpendicular to dir **d** of r(t)

(PQ)X(PX)•**d** = 0, or, in determinant form: |(PQ);(PX);**d**|=0

(PQ)=(3,2,-1) - (1,-1,2)=(2,3,-1)

(PX)=(3,2,-1) – (x,y,z)=[(3-x),(2-y),(-1-z)]

**d**=(3,2,-1)

3x-7y-5z=0

Re: Equation of plane through two points and parallel to line

Quote:

Originally Posted by

**OVOXO** So, I take PQ X R as my coefficients of the plane?

(-2,-3,3)x(3,2,-1)=(-3,7,5)

-3x+7y+5z=0 is parallel to the line and includes p and q?

How do you know that d = 0? Use a point you know lies on the plane as (x, y, z) in the equation of your plane to evaluate d.

Re: Equation of plane through two points and parallel to line

Thanks Prove It, you are right:

**l** = (3.2,-1) dir of line

PQX**l** = (2,3,-3)X(3,2,-1) = (3,-7,-5) is normal to plane par to line

3x-7y-5x=d is equation of plane.

Substitute P = (3,2,-1) to determine d.

Re: Equation of plane through two points and parallel to line

I had subbed it in and d was equal to zero.

Re: Equation of plane through two points and parallel to line

Good. That was what Prove It was asking.