Finding if a line intersects the plane or not

**Giestforlife** · February 7th 2013, 02:48 PM

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.

**Plato** · February 7th 2013, 03:04 PM

Originally Posted by Giestforlife

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!

Given a line $\ell: P+tD$ and plane $\Pi: N\cdot<x-a,y-b,z-c>=0$ then $\ell\|\Pi$ if and only if $D \cdot N=0.$ .
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.

**Giestforlife** · February 7th 2013, 06:30 PM

Originally Posted by Plato

Given a line $\ell: P+tD$ and plane $\Pi: N\cdot<x-a,y-b,z-c>=0$ then $\ell\|\Pi$ if and only if $D \cdot N=0.$ .
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.

Damn, I guess I just got tunnel visioned. But thanks so much!

**Soroban** · February 7th 2013, 07:07 PM

Hello, Giestforlife

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

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

and we get: . $11 \,=\,5\,??$ . . . 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:. $\vec v \,=\,\langle\text{-}4,0,2\rangle$
The normal vector of the plane is:. $\vec n \,=\,\langle 1,2,2\rangle$

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

The line and the normal vector are perpendicular.
Therefore, the line and plane are parallel.

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