The direction of the line is perpendicular to the plane, so the direction of the line is parallel to the plane's normal.Kawrider0133 said:Point: (2,3,-5)
Perpendicular to the line: x = -1 - t, y = 4 + 3t, z = 4t
I've looked up many examples to similar problems but i just can't get the right answer to this one. Please Help!
The equation of a plane that passes through a given point (x1, y1, z1) and is perpendicular to a given line with direction vector (a, b, c) can be written as: ax + by + cz = d, where d = ax1 + by1 + cz1.
To find the equation of a plane given its normal vector (a, b, c) and a point (x1, y1, z1), you can use the formula: a(x-x1) + b(y-y1) + c(z-z1) = 0.
No, the equation of a plane requires at least three points to be determined. If only two points are given, there are infinite planes that can pass through them.
The distance between a point and a plane affects the value of d in the equation of the plane (ax + by + cz = d). The farther away the point is from the plane, the larger the value of d will be.
The normal vector of a plane is a vector that is perpendicular to the plane. It is represented by (a, b, c) in the equation of the plane (ax + by + cz = d). The coefficients a, b, and c represent the direction of the normal vector and can be used to find the angle between the plane and another line or plane.