Final answer:
The student's question involves finding the vector form of a line in three-dimensional space that is perpendicular to a given plane. Using the plane's normal vector as the direction vector, we can express the vector form of the line.
Step-by-step explanation:
The student is asking to find the vector form of an equation of a line in three-dimensional space (R^3) that passes through a specific point and is perpendicular to a given plane. To do this, we can use the normal vector of the plane to serve as the direction vector for the line, since any line perpendicular to a plane will be parallel to the normal vector of the plane.
The general equation of the plane is given as x - 3y + 2z - 5. The coefficients of x, y, and z in the equation of the plane give us the components of the normal vector, which are (1, -3, 2). Therefore, the vector form of the line can be expressed by r(t) = P + t * N, where P is the point the line passes through, N is the normal vector, and t is a scalar parameter. Substituting in the given point and the normal vector we have r(t) = (-1, 0, 3) + t * (1, -3, 2).
To find the vector form of the equation of the line that passes through point P(-1, 0, 3) and is perpendicular to the plane with the equation x - 3y + 2z - 5 = 0, we need to find a vector that is perpendicular to the plane. The coefficients of x, y, and z in the plane equation represent the direction of the normal vector to the plane. Therefore, the normal vector is (1, -3, 2).