find the line that passes through the point (2,5,3) and is perpendicular to the plane 2x-3y+4z+7 = 0
this problem is almost identical to the last, so I'm gonna make you do a lot more on your own this time ;) 1) what is \(\vec n\) of the given plane?
(2,-3,4) ?
yes, nice I just realized that this problem is a bit different though, because it asks for the equation of a normal line, not for the plane similar business though; we still needed the normal vector now we need a line that is parallel to that vector I don't suppose that you know that the equation of a line through a point \(P_0\) is given by\[\vec r(t)=\vec r_0+t\vec v\]where \(\vec r_0\) is the position vector of the given point, and \(\vec v\) is a vector parallel to the line, but that is supposed to be know to solve this problem. I will try to help you understand it anyway though...
here is a link on the ideas I will express in my explanation http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx please open it and look at the picture of the red line
in this case to use our formula we need to not that the parallel vector in the line formula \(\vec v=\vec n\) because we want a line perpendicular to the plane, which implies it must be parallel to \(\vec n\) because of that we can rewrite this as\[\vec r=\vec r_0+t\vec n\]now you already know \(\vec n\), so now tell me what is \(\vec r_0\) (the position vector that points to the given point)
(2,5,3)?
yep, so now you can just plug int the stuff we know why don't you give that a try?
= (2,5,3)+(2,-3,4) ?
don't forget the t, which we treat with scalar multiplication (scalar multiplication is this:\(c\vec v=t<a,b,c>=<ta,tb,tc>\))\[\vec r(t)=\vec r_0+t\vec n\]\[\vec r(t)=<2,5,3>+t<2,-3,4>\]\[\vec r(t)=<2+2t,5-3t,3+4t>\]
thanks
getting it slowly
by looking at the individual components we can also write this as\[x=2+2t\]\[y=5-3t\]\[z=3+4t\]there are a number of ways to write the equation of a line
thanks.. next question?
give me a few minuts, I am eating breakfast and my cat just broke something :(
haha ok no prob, im goin to take a small break, i will post a new thread for numbr 3, plz reply wen u available..
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