Find the equation in the form of ax+by+cz=d

  • Thread starter XBOX999
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  • #1
XBOX999
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Hello,
I need help in this two quetion. Please, show me all the steps to solve them.

1- Find the distence from a point P= ( -3,4,2) to a line L:(x,y,z)= ( 3,-2,-1)+(1,-2,2)t.



2-find the equation in the form of ax+by+cz=d, of the plane going through a point
P= (-3,4,2) to a line L:(x,y,z)= (3,-2,-1) + (1,-2,2)t.
 

Answers and Replies

  • #2
jquark
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i'm at the library, i was just looking at those in my crc standard formulas and tables around page 240-250. its a piece of cake but i cant quote them. someone will show up.
 
  • #3
jbunniii
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Hello,
I need help in this two quetion. Please, show me all the steps to solve them.

1- Find the distence from a point P= ( -3,4,2) to a line L:(x,y,z)= ( 3,-2,-1)+(1,-2,2)t.



2-find the equation in the form of ax+by+cz=d, of the plane going through a point
P= (-3,4,2) to a line L:(x,y,z)= (3,-2,-1) + (1,-2,2)t.

For the first part, the distance from a point to a line is by definition the SHORTEST possible distance along a line connecting P and L. At what angle must the two lines meet in order for the distance to be minimized?

In other words, if q is a point on the line, you want ||q - p|| to be the shortest possible. If v is a vector pointing in the direction of the line, what condition must the vectors (q - p) and v satisfy?
 

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