Find the equation of the plane, given parametric equations
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Find the equation of the plane, given parametric equations

[From: ] [author: ] [Date: 11-04-22] [Hit: ]
4, -1) ...Vector from (-1, 5,......
The parametric equations:

x = -1 + 3t
y = 5 + 2t
z = 2 + t

and passes thru (2, 4 ,-1)

what is the equation of the plane????

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We need to find two vectors that are parallel to the plane.

First vector is direction vector of line: < 3, 2, 1 >

We find two points on plane:
(-1, 5, 2) which is on the given line
(2, 4, -1) ... given

Vector from (-1, 5, 2) to (2, 4, -1) = < 3, -1, -3 >

Two vectors parallel to plane are < 3, -1, -3 > and < 3, 2, 1 >
Cross product of these two vectors is normal to the plane
< 3, -1, -3 > x < 3, 2, 1 > = < 5, -12, 9 >

Plane has normal vector < 5, -12, 9 > and passes through point (2, 4, -1)

Equation of plane:
5 (x - 2) - 12 (y - 4) + 9 (z + 1) = 0
5x - 10 - 12y + 48 + 9z + 9 = 0
5x - 12y + 9z + 47 = 0

5x - 12y + 9z = -47
1
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