Does the line contain in the plane?

**Jenny20** · January 5th 2007, 03:44 PM

Give x=3-t, y=2+t, z=1-3t
2x+2y-5=0

a) determine whether the line and plane are parallel, perpendicular , or either.
My answer to part a) is : the line is parallel to the given plane.

b) Does the line contain in the plane?
How should I answer this part?

Thank you very much.

**Plato** · January 5th 2007, 04:19 PM

Does $2\left( {3 - t} \right) + 2\left( {2 + t} \right) - 5 = 0$ for every t?
If so the line is in the plane.

**Soroban** · January 5th 2007, 06:02 PM

Hello, Jenny!

Given: . $\begin{array}{ccc}x\:=\:3-t \\ y\:=\:2+t \\ z\:=\:1-3t\end{array},\;\;2x + 2y - 5 \:= \:0$

(a) Determine whether the line and plane are parallel, perpendicular, or neither.
My answer to part (a) is: the line is parallel to the given plane . . . yes

(b) Is the line contained in the plane?

(a) The line has direction vector: $\vec{v} \:=\:\langle -1,1,-3\rangle$
The plane has normal vector: $\vec{n} \:=\:\langle 2,2,0\rangle$

If the line and plane are perpendicular, then $\vec{v} = k\vec{n}$ for some nonzero number $k.$
. . This is not true . . . they are not perpendicular.

If the line and plane are parallel, then: $\vec{v}\cdot\vec{n} = 0$
. . Since $\langle -1,1,-3\rangle\cdot\langle2,2,0\rangle \:=\;0$ . . . they are parallel.

(b) The equation of the plane is: . . . . . . $2x \quad+ \quad2y - 5 \:=\:0$
Substitute the parametric equations: . $2\overbrace{(3 - t)}^\downarrow + 2\overbrace{(2 + t)}^\downarrow - 5 \:=\:0$

If this statement is always true, the line lies in the plane.

We get: . $6 - 2t + 4 + 2t - 5 \:=\:0\quad\Rightarrow\quad 5 \,=\,0\:??$
. . Hence, the line does not lie in the plane.

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