Find equation of the plane that passes through a given point and is perpen. to

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To find the equation of the plane that passes through the point (2,3,-5) and is perpendicular to the given line, first identify the direction vector of the line, which is derived from its parametric equations. The direction vector is (-1, 3, 4), and this vector serves as the normal vector for the plane. Using the point-normal form of the plane equation, the equation can be formulated as -1(x - 2) + 3(y - 3) + 4(z + 5) = 0. Understanding how to derive the normal vector from the line's direction and applying it with the point on the plane is crucial for solving this problem. The correct application of these concepts will yield the desired plane equation.
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Find equation of the plane that passes through a given point and is perpen. to...

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!
 
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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 direction of the line is perpendicular to the plane, so the direction of the line is parallel to the plane's normal.

Do you know how to find a vector in the direction of a given line?
Do you know how to find the equation of a plane given a point on the plane and the normal?
 
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