Gradient scalar field Watch

mathslover786
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#1
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#1
thanks
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nuodai
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#2
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Nope... how have you worked out your partial derivatives? Post your working and I'll show you where you've gone wrong.
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mathslover786
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(Original post by nuodai)
Nope... how have you worked out your partial derivatives? Post your working and I'll show you where you've gone wrong.
think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?
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nuodai
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(Original post by mathslover786)
think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?
Yup
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mathslover786
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(Original post by nuodai)
Yup
Thank You!
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mathslover786
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#6
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(Original post by nuodai)
Yup
another quick question:


Show that the following points form a right angled triangle :

A( 1 , 2 , 3 ) , B( 3 , 8 , 13 ) and C( –2 , 8 , 14 ) .

Also, find the area of the triangle, and the equation of the plane on which the triangle sits .

cos(theta) between two of the letters has to be 1 i know but it doesnt seem to work...HELP!
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nuodai
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(Original post by mathslover786)
another quick question:


Show that the following points form a right angled triangle :

A( 1 , 2 , 3 ) , B( 3 , 8 , 13 ) and C( –2 , 8 , 14 ) .

Also, find the area of the triangle, and the equation of the plane on which the triangle sits .

cos(theta) between two of the letters has to be 1 i know but it doesnt seem to work...HELP!
There's your error. The letters are the position vectors of the three vertices of the triangle, not the sides. The sides are \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{BC}, and it's two of these that have to be perpendicular for it to be a right-angled triangle.
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mathslover786
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#8
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(Original post by nuodai)
There's your error. The letters are the position vectors of the three vertices of the triangle, not the sides. The sides are \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{BC}, and it's two of these that have to be perpendicular for it to be a right-angled triangle.
ahhhhh yeah i get it now lol...i was using them as the length of the sides do you have your own youtube channel...it would be verry useful
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mathslover786
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#9
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(Original post by mathslover786)
think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?
for the Curl of this i got 0 is this correct?
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