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.
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.
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.
Hello, Giestforlife
$\displaystyle \text{Does the line }\begin{Bmatrix}x &=& 1-4t \\ y&=& 3 \\ z &=& 2t +2\end{Bmatrix}\,\text{ intersect the plane: }\,x + 2y +2z \:=\:5\,?$
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 does not intersect the plane.
. . 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.