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????
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
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