# Math Help - equation of line and plane !! plzzzz help

asd

2. ## Re: equation of line and plane !! plzzzz help

Originally Posted by alakd
Can someone help with this multi-part question. i did the first three but it doesnt seem right!
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(a) Find the equation of the line l through P(1, 1, 2) and Q(1, 0, 4) in vector, parametric and Cartesian forms.
(b) Find the vector form of the line k through R(0, 3, 1) which is perpendicular to l.
(c) Find the Cartesian form of the plane containing the lines l and k.
(d) Find the vector form of the line through T(0, 0, 3) which is normal to this plane.
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so i guess i did the first bit alright : cartesian : z=6+4x , y=6
for the second one i got this line : (-5,4,2)+t(28,-37,7) , i timed it by 17 to get rid of the denominator
third part : the lines seem to have no intersection so i am stuck!

appreciate it if anyone can help out
The line will be infinitely long and in the direction of the vector that goes from \displaystyle \begin{align*} \overline{OP} \end{align*} to \displaystyle \begin{align*} \overline{OQ} \end{align*}. So

\displaystyle \begin{align*} \overline{PQ} &= \overline{PO} + \overline{OQ} \\ &= -\overline{OP} + \overline{OQ} \\ &= \overline{OQ} - \overline{OP} \\ &= <1, 0, 4> - <1, 1, 2> \\ &= <0, -1, 2> \end{align*}

To make it infinitely long, we multiply by some parameter, \displaystyle \begin{align*} t \end{align*}, to give \displaystyle \begin{align*} t<0, -1, 2> \end{align*}.

Then we need to position this infinitely long vector at a point. Adding either of the points you know the line goes through will do.

So \displaystyle \begin{align*} l = <1, 1, 2> + t<0, -1, 2> \end{align*}.

This is the vector form of the line.

To get the parametric equations, expand it out...

\displaystyle \begin{align*} l &= <1, 1 - t, 2 + 2t> \\ \\ x &= 1 \\ \\ y &= 1 - t \\ \\ z &= 2 + 2t \end{align*}

To get the Cartesian equations, try to write each of them as t in terms of the other variable and set them equal.