Cartesian Equation of a Plane

coma

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Apr 15, 2021
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"Write the cartesian equation of the plane containing the line π‘Ÿβƒ—1 = (1, 2, 4) + 𝑑(4, 1, 11) and perpendicular to π‘Ÿβƒ—2 = (4, 15, 8) + 𝑠(2, 3, βˆ’1)"

Haven't been able to figure out where to even begin with this one, visualizing it in my head makes sense but I've got no idea how to solve it algebraically. It's worth 3 marks if that's helpful.
 

Dr.Peterson

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"Write the cartesian equation of the plane containing the line π‘Ÿβƒ—1 = (1, 2, 4) + 𝑑(4, 1, 11) and perpendicular to π‘Ÿβƒ—2 = (4, 15, 8) + 𝑠(2, 3, βˆ’1)"

Haven't been able to figure out where to even begin with this one, visualizing it in my head makes sense but I've got no idea how to solve it algebraically. It's worth 3 marks if that's helpful.
What forms have you learned for the equation of a plane?

You have a normal vector, and a lot of points in the plane (a whole line's worth), so you should have an appropriate form for the equation. (In fact, you could have too much information -- you'll need to check that the first line actually lies in this plane. That may be what is confusing you.)
 

pka

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"Write the cartesian equation of the plane containing the line π‘Ÿβƒ—1 = (1, 2, 4) + 𝑑(4, 1, 11) and perpendicular to π‘Ÿβƒ—2 = (4, 15, 8) + 𝑠(2, 3, βˆ’1)"
Let's rewrite the notation.
\({\ell _1}(t) = \left\{ \begin{gathered} x = 1 + 4t \hfill \\ y = 2 + t \hfill \\ z = 4 + 11t \hfill \\ \end{gathered} \right.~~\) \({\ell _2}(s) = \left\{ \begin{gathered} x = 4 + 2s \hfill \\ y = 15 + 3s \hfill \\ z = 8-s \hfill \\ \end{gathered} \right.\)
Now we need the point \((1,2,4)\) and the normal vector \(\vec{n}=\left<2,3,-1\right>\)
 

pka

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Let's rewrite the notation.
\({\ell _1}(t) = \left\{ \begin{gathered} x = 1 + 4t \hfill \\ y = 2 + t \hfill \\ z = 4 + 11t \hfill \\ \end{gathered} \right.~~\) \({\ell _2}(s) = \left\{ \begin{gathered} x = 4 + 2s \hfill \\ y = 15 + 3s \hfill \\ z = 8-s \hfill \\ \end{gathered} \right.\)
Now we need the point \((1,2,4)\) and the normal vector \(\vec{n}=\left<2,3,-1\right>\)
Edit P. S. correction of a miss-copy.
\(\vec{n}=\left<4,1,11\right>\times\left<2,3,-1\right>=\left<-17,3,5\right>\)
 

pka

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Edit P. S. correction of a miss-copy.
\(\vec{n}=\left<4,1,11\right>\times\left<2,3,-1\right>=\left<-17,3,5\right>\)
A second correction.
\(\vec{n}=\left<4,1,11\right>\times\left<2,3,-1\right>=\left<-17,13,5\right>\)
 

lex

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@pka
Your original post is correct.

@coma
You can write down the equation of a plane if you know a point in the plane and a vector normal to the plane. You can read off a point from the first equation and read off a normal vector from the direction vector in the second equation, as indicated by @pka in his original post.
 

lex

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(To see that such a plane exists, we note that the direction vectors of the two lines are perpendicular to each other (4,1,11), (2,3,-1))
 
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