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- 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:

P

Therefore, the answer is [6,0,0] +

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

My answer is

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

**u**= [-2,3,1]P

_{o}= (6,0,0) & P = (4,2,3)**P**[-2,2,3]_{o}P = v =Therefore, the answer is [6,0,0] +

**r**[-2,3,1] +**q**[-2,2,3]; r, q are real numbersI 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 numbersCould anyone explain the solution, and is my answer correct? Thanks.