## orthogonal

Find an equation of a plane through the point (-1, -2, -3) which is orthogonal to the line x=5+2t,y=-3-5t,z=2-2t
in which the coefficient of x is 2.

______________________________=0

i dont get this problem at all, but here's what i came up with after sitting here at the computer for a long time attempting to do this problem.

okay i know that by theorem, The vector a X b is orthogonal to both a and b.

a = <-1,-2,-3>
b= <2,-5,2>

that's all i came up, i attempted many different ways, but it doesnt make sense at all, can someone lend me a hand?
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 Recognitions: Gold Member Homework Help Science Advisor Ok, here's a plan: Find out the general form of a vector parallel to the line. (it will be a function of two parameters; say you labeled them t_1 and t_2.)Write out the form of a vector that lies in the plane (it will be a function of x,y & z). Then take the dot product of those two vectors and set it equal to 0. This is the equation of your plane. Now chose a particular vector parrarel to the line such that the coeff. of x is 2... that is to say, set t_1 and t_2 such that the coeff. of x is 2. I understand that this might sound very confusing but try your best to progress. If you get stuck, or if after thinking very hard about it, you still can't find a way, I'll help you some more. Good luck!!
 N*r = n*r_0 is the equation im looking for right? i am totally lost here, book doesnt help much either, can you help me start it.

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Let's start with the first step of my plan. Let's find the general form of a vector parallel to the line. The equation of the line is written in parametric form. It means that for any value of t, the point (x,y,z)=(5+2t,-3-5t,-2-2t) is on the line. Ok, so let's find two points on the line by setting t = t_1 and t = t_2. Our two points are $\vec{P}_1 = (5+2t_1,-3-5t_1,-2-2t_1)$ and $\vec{P}_2 = (5+2t_2,-3-5t_2,-2-2t_2)$. By means of a simple drawing, you can convince yourself that $\vec{P}_1 - \vec{P}_2 = (2(t_1-t_2), -5(t_1-t_2), -2(t_1-t_2))$ is a vector parallel to the line (it is in fact IN/MERGED WITH the line).