If the line x = 1-4t; y = 3; z = 2t +2 intersects the plane x + 2y +2z = 5?

I already tried plugging in the parametrization of the line into the equation of the plane, but I get an indeterminate answer!

Any help is greatly appreciated.

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- Feb 7th 2013, 02:48 PMGiestforlifeFinding if a line intersects the plane or not
If the line x = 1-4t; y = 3; z = 2t +2 intersects the plane x + 2y +2z = 5?

I already tried plugging in the parametrization of the line into the equation of the plane, but I get an indeterminate answer!

Any help is greatly appreciated. - Feb 7th 2013, 03:04 PMPlatoRe: Finding if a line intersects the plane or not
Given a line $\displaystyle \ell: P+tD$ and plane $\displaystyle \Pi: N\cdot<x-a,y-b,z-c>=0$ then $\displaystyle \ell\|\Pi$ if and only if $\displaystyle D \cdot N=0.$.

That is the direction of the line is perpendicular to the normal of the plane.

If a line is not parallel to a plane then they intersect. - Feb 7th 2013, 06:30 PMGiestforlifeRe: Finding if a line intersects the plane or not
- Feb 7th 2013, 07:07 PMSorobanRe: Finding if a line intersects the plane or not
Hello, Giestforlife

Quote:

$\displaystyle \text{Does the line }\begin{Bmatrix}x &=& 1-4t \\ y&=& 3 \\ z &=& 2t +2\end{Bmatrix}\,\text{ intersect the plane: }\,x + 2y +2z \:=\:5\,?$

I tried plugging in the parametrization of the line into the equation of the plane,

. . but I get an indeterminate answer!

Not sure what you mean by "indeterminate".

We have: .$\displaystyle (1-4t) + 2(3) + 2(2t+2) \:=\:5 \quad\Rightarrow\quad 1 - 4t + 6 + 4 + 4t \:=\:5$

and we get: . $\displaystyle 11 \,=\,5\,??$ . . . a false statement.

This means that the line doesintersect the plane.*not*

. . They are parallel.

We can check this fact.

The direction vector of the line is:.$\displaystyle \vec v \,=\,\langle\text{-}4,0,2\rangle$

The normal vector of the plane is:.$\displaystyle \vec n \,=\,\langle 1,2,2\rangle$

And:.$\displaystyle \vec v \cdot \vec n \:=\:\langle\text{-}4,0,2\rangle\cdot\langle1,2,2\rangle \:=\:\text{-}4 + 0 + 4 \:=\:0$

The line and the normal vector are perpendicular.

Therefore, the line and plane are parallel.