Finding the vector equation of a plane

  • #1
Homework Statement:
Find a vector equation of the plane that passes through the point (6, 0, 0) and contains the line 𝑥 = 4 - 2𝑡, 𝑦 = 2 + 3𝑡, 𝑧 = 3 + 𝑡.
Relevant Equations:
vector equation
Solution:
u = [-2,3,1]
Po = (6,0,0) & P = (4,2,3)
PoP = v = [-2,2,3]
Therefore, the answer is [6,0,0] + r[-2,3,1] + q[-2,2,3]; r, q are real numbers

I don't understand why (6,0,0) is used as the point in the vector equation, since it only lies on the [-2,2,3] vector, not the u = [-2,3,1] vector.

My answer is r = [4,2,3] + r[2,-2,-3] + q[-2,3,1], r, q are real numbers

Could anyone explain the solution, and is my answer correct? Thanks.
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,025
467
Your answer is transformable to
[tex][4,2,3]+q[-2,3,1]-r[-2,2,3]=[6,0,0]+q[-2,3,1]-(r-1)[-2,2,3][/tex]
changing q with r and -(r-1) with q, it equals to the text answer.
 
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