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Find the equation of the plane which contains the point (3,2,3) and the line (x,y,z) 
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#1
Nov1709, 09:11 PM

P: 103

1. The problem statement, all variables and given/known data
Find the equation of the plane which contains the point (3,2,3) and the line: (x,y,z) = (7,4,5) + t (0,2,2) 2. Relevant equations PointNormal equation? a(xx_{0})+b(yy_{0})+c(zz_{0}) = 0 Not sure if this is related. 3. The attempt at a solution First off: The line (xyz) should be (7,4,5) + t (0,2,2) which becomes (7, 42t, 5 +7t). And then yea... I'm behind :(. I've been sick for the past 2 weeks and I don't understand my friends' notes. So I'm not asking for any of you to solve this for me, but if someone could point me in the right direction and help me on interpreting the question, then that would be great. Thanks. 


#2
Nov1709, 09:18 PM

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You can use the normal equation once you find a normal. You've got one direction vector which is parallel to the plane which is the direction vector to the line. Find another one by taking the difference between your point and any point on the line. Then use the cross product.



#3
Nov1709, 09:20 PM

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Or just take two different values of t to find two different points on the line. You now have three points in the plane. You know how to find the plane containing three given points, don't you?



#4
Nov1709, 10:04 PM

P: 103

Find the equation of the plane which contains the point (3,2,3) and the line (x,y,z)
ax + by + cz + d = 0 to get the equation of each point and solve for the system of equations? Is there a difference between "passing through the points" and "contains the points" in terms of planes? 


#5
Nov1709, 10:14 PM

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#6
Nov1809, 05:22 AM

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But the way most people learn to find the plane containing three given points is to determine the vectors from one of the points to the other two and take the cross product of those which leads back to the method Dick is suggesting. 


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