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???
You can do a bit more than that. If $\displaystyle (x_0, y_0, 7)$ lies on the plane 4x+ y- 3z= 1 then [tex]4x_0+ y_0- 21= 1 so that $\displaystyle y_0= 22- x_0$ and you can write the line as $\displaystyle x= x_0+ 4t$, $\displaystyle y= 22- x_0+ t$, and $\displaystyle z= 7- 3t$.
Hello, rikelda91!
This is a strange problem.
Do you have the answer?
$\displaystyle \text{Find the line that is vertical to the plane }4x+y-3z\:=\:1$
. . $\displaystyle \text{and passes through the point }A(x_o,y_o,7)\text{ of the plane.}$
$\displaystyle \text{The direction of the line is the normal to the plane: }\,\vec n \,=\,\langle 4,1,\text{-}3\rangle$
$\displaystyle \text{Since point }A\text{ is on the plane: }\:4x_o + y_o - 3(7) \:=\:1 \quad\Rightarrow\quad y_o \,=\,22-4x_o$
$\displaystyle \text{We have point }A(x_o,\,22\!-\!4x_o,\,7)\,\text{ and vector }\,\vec n \,=\,\langle 4,1,\text{-}3\rangle$
$\displaystyle \text{The line is: }\:\begin{Bmatrix}x &=& x_o + 4t \\ y &=& (22\!-\!4x_o) + t \\ z &=& 7 - 3t \end{Bmatrix}$
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???