Lines and planes (Scalar equation + pt. of int.)

[Lornt]

so I've been doing review and I got stuck on two questions, I was wondering if anyone could help me out.
Any help would be appericiated - whether it's a hint on what my first step should be or what's the final answer to confirm whether I'm right or wrong


1. Find the scalar equation of the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the planes x + 2y + z = 1 and 2x + y + 3z = 3.


2. Determine the distance from the point P(2, 3, -1) to the plane 2x + y - 2z + 9 = 0. (I know I can plug the values into the equation, but that doesn't really help me. Could anyone show me a longer way to do this involving more steps? I don't really understand where the equation comes from).

 

[galactus]

1. Find the scalar equation of the plane that passes through the
point (1, 1, 4) and is perpendicular to the line of intersection of the planes
x + 2y + z = 1 and 2x + y + 3z = 3.

\(\displaystyle n_{1}=<1,2,1>; n_{2}=<2,1,3>\) are normals to the given

planes. CrossP= \(\displaystyle n_{1}\times{n_{2}}=<5,-1,-3>\)

So, \(\displaystyle <-5,1,3>\) is normal to the desired plane whose equation

is \(\displaystyle \L\\-5(x-1)+(y-1)+3(z-4)=-5x+y+3z-8=0\)

For #2, use the formula for the distance berween a point and a plane:

\(\displaystyle \L\\D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

 

[soroban]

Hello, Lornt!

2. Determine the distance from the point \(\displaystyle P(2, 3, -1)\) to the plane \(\displaystyle \,2x\,+\,y\,-\,2z\,+\,9\:=\:0\)

(I know I can plug the values into the formula, but that doesn't really help me.
Could anyone show me a longer way to do this involving more steps?
I don't really understand where the formula comes from).

We want the equations of the line \(\displaystyle L\) through \(\displaystyle P\), perpendicular to the plane.

The plane has the normal vector: \(\displaystyle \,\vec{n}\:=\:\langle 2,1,-2\rangle\) . . . and that is the direction of line \(\displaystyle L\).

Hence, line \(\displaystyle L\) has equations: \(\displaystyle \,\begin{Bmatrix}x\:=\:2\,+\,2t \\ y\:=\:3\,+\,t \\ z\:=\:-1\,-\,2t\end{Bmatrix}\)


To find the intersection \(\displaystyle Q\) of line\(\displaystyle L\) and the plane,
\(\displaystyle \;\;\)substitute the parametric equations into the equation of the plane:

\(\displaystyle 2(2\,+\,2t)\,+\,(3\,+\,t)\,-\,2(-1\,-\,2t)\,+\,9\;=\;0\;\;\Rightarrow\;\;9t\,+\,18\:=\:0\;\;\Rightarrow\;\;t\,=\,-2\)


Hence: \(\displaystyle \begin{array}{ccc}x\:=\:2\.+\.2(-2)\:=\:-2\\ y\:=\:3\,+\,(-2)\:=\:1 \\ z\:=\:-1\,+\,2(-2)\:=\:3\end{array}\;\;\Rightarrow\;\;Q(-2,1,3)\)


The distance from \(\displaystyle P(2,3,-1)\) to \(\displaystyle Q(-2,1,3)\) is given by:

\(\displaystyle \;\;\overline{PQ} \:=\:\sqrt{(-2\,-\,2)^2\,+\,(1\,-\,3)^2\,+\,(3\,+\,1)^2} \:=\:\sqrt{36}\:=\:6\)