Answer: We can find the equation of a plane in point-normal form, which is given by:
Ax + By + Cz + D = 0
where (A, B, C) is the normal vector to the plane and (x, y, z) are the coordinates of any point on the plane, and D is a constant.
To find the normal vector, we can use the cross product of two vectors that lie on the plane. A convenient choice for these vectors is the direction vector of the line, which is given by:
v = <2-3, -1-4, 5-(-1)> = < -1, -5, 6 >
The normal vector to the plane is given by the cross product of the direction vector of the line and the vector pointing from any point on the line to the point (2,-3,1). Let's call this vector v2 = <2-3, -3-4, 1-(-1)> = < -1, -7, 2>
So the normal vector is given by:
n = v x v2 = ( -52 + 6-7)i + (6*-1 - -12)j + (-1-7 - -5*2)k = (14)i + (-7)j + (-11)k
Given that the point (2,-3,1) lies on the plane, we can substitute these values in the point-normal form equation:
A2 + B-3 + C*1 + D = 0
-142 + -7-3 + -11*1 + D = 0
-28 - 21 - 11 + D = 0
D = 58
So the equation of the plane is :
14x - 7y - 11z + 58 = 0
This is the equation that represents the plane that passes through (2,−3,1) and is normal to the line joining the points (3,4,−1) and (2,−1,5)
Explanation: