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Write the Cartesian equation for the plane containing the point (2,-1,8) and perpendicular to the line [x,y,z] = [1,-2,-3] + s[5,-4,7].
The situation is that I have my Calc. + Vectors exam tomorrow morning and I'm just going through some questions on my old tests. This one has got me completely lost.
Any help to get me started would be appreciated.
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In addition, I have a second question that I thought I got correct but it was marked as incorrect so if anyone can tell me what I did wrong, it would be great.
Sketch the plan (was it a typo and not plane?) given by 3x+4y-10=0
I found the x-int to be 10/3 and y-int to be 2.5. I simply connected the two points. What did I do wrong?
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With some help from others, I was able to get the answer. In case anyone in the future is interested:
Since the line is perpendicular to the plane and we know that the normal of the plane is also perpendicular to it, we can conclude that the normal of the plane and the direction vector of the line are parallel.
So we can use Ax+By+Cz+D=0 where A=5, B=-4, C=7. x,y,z are simply coordinates of the point in the plane (2,-1,8)
(5)(2)+(-4)(-1)+(7)(8)+D=0
D=-70
Therefore, the Cartesian equation of the plane is 5x-4y+7z-70=0.
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Hi Anakin;
Haven't seen you for a while.
Sketch the plan (was it a typo and not plane?) given by 3x+4y-10=0
I found the x-int to be 10/3 and y-int to be 2.5. I simply connected the two points. What did I do wrong?
Aren't you forgetting the z coordinate? I have in my notes ( 10 / 3, 0, z ) and (0, 10 / 4, z ) where z is an arbitrary parameter.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hey Bobby, yeah I've been doing fairly well in the Calculus part (got a head start from the derivative lesson you gave here a long time back) so never gave much attention to it, hence my absence.
As for the question, if one subbed in 0 for z then in the two coordinates, would the line simply look like one that connects the x-int and y-ints?
I really should be knowing this. My exam's tomorrow and I practically just started vectors this weekend.
Last edited by Anakin (2010-06-24 18:57:33)
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Hi Anakin;
I think so but I am not sure. But you have a family of points not just z = 0. You could have z = 1 and z = -9.3 etc.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yup. I was just putting z=0 for ease but since you mentioned we can put any number, does that mean the plane can have more than one sketch?
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Well, if we vary z in (x, y, z) keeping x , y constant arent we drawing a vertical line?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I wouldn't know so I'll take your word for it.
So by changing the z coordinate, are we just shifting the line across the z-axis and nothing else?
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Yes, I believe so, like holding x constant and varying y creates a vertical line. Same thing here.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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That makes good sense. Thanks once again Bobbym.
Wish me luck on the exam, it's the last thing to worry about before my long summer.
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Hi Anakin;
The best of luck. Stay calm, you will do well. Bring what you have a problem with back here we will try figure it out for next time.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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