# Find the line

• August 20th 2013, 04:40 AM
rikelda91
Find the line
Find the line that is vertical to the plane 4x+y-3z=1 and passes through the point A(x0,y0,z0) of the plane so that z0=7.
Could anyone help me with that exercise???
• August 20th 2013, 07:07 AM
Plato
Re: Find the line
Quote:

Originally Posted by rikelda91
Find the line that is vertical to the plane 4x+y-3z=1 and passes through the point A(x0,y0,z0) of the plane so that z0=7.

I think that there is a translation problem here.
By vertical to the plane you surely mean perpendicular to the plane. Do you not?
And the line contains $(x_0,y_0,7)~?$

Such a line looks like $\\x=x_0+4t\\y=y_0+t\\z=7-3t$
• August 20th 2013, 07:25 AM
HallsofIvy
Re: Find the line
You can do a bit more than that. If $(x_0, y_0, 7)$ lies on the plane 4x+ y- 3z= 1 then [tex]4x_0+ y_0- 21= 1 so that $y_0= 22- x_0$ and you can write the line as $x= x_0+ 4t$, $y= 22- x_0+ t$, and $z= 7- 3t$.
• August 20th 2013, 07:29 AM
Soroban
Re: Find the line
Hello, rikelda91!

This is a strange problem.

Quote:

$\text{Find the line that is vertical to the plane }4x+y-3z\:=\:1$
. . $\text{and passes through the point }A(x_o,y_o,7)\text{ of the plane.}$

$\text{The direction of the line is the normal to the plane: }\,\vec n \,=\,\langle 4,1,\text{-}3\rangle$

$\text{Since point }A\text{ is on the plane: }\:4x_o + y_o - 3(7) \:=\:1 \quad\Rightarrow\quad y_o \,=\,22-4x_o$

$\text{We have point }A(x_o,\,22\!-\!4x_o,\,7)\,\text{ and vector }\,\vec n \,=\,\langle 4,1,\text{-}3\rangle$

$\text{The line is: }\:\begin{Bmatrix}x &=& x_o + 4t \\ y &=& (22\!-\!4x_o) + t \\ z &=& 7 - 3t \end{Bmatrix}$
• August 20th 2013, 07:50 AM
rikelda91
Re: Find the line
no I don't have the answer :/ ..actually I meant perpendicular to the plane...
the line equation should be l(t)=a+tu (u is a parallel vector to the line). So l(t)=(x0,22-4x0,7) +t(4,1,-3)
Right???