Final answer:
The vector equation for the line that is perpendicular to the plane x + 2y + z = 3 and passes through the point (1, 0, 3) is r(t) = (1 + t)i + (2t)j + (3 + t)k, and the parametric equations are x = 1 + t, y = 2t, z = 3 + t.
Step-by-step explanation:
To find both a vector equation and parametric equations for the line that passes through the point (1, 0, 3) and is perpendicular to the plane x + 2y + z = 3, we must first find the normal vector to the plane, which is also the direction vector for the line. This normal vector can be directly read from the coefficients of x, y, and z in the plane's equation, which are (1, 2, 1).
Next, we use the given point (1, 0, 3) as the point through which the line passes. The vector equation of the line can be represented as r(t) = r0 + tv, where r0 is the position vector to the point (1, 0, 3), and v is the direction vector (1, 2, 1). Writing the position vector and multiplying the direction vector by the parameter t, we get the following form:
r(t) = (1 + t)i + (2t)j + (3 + t)k.
The parametric equations are derived from the vector equation by equating each component to its corresponding parameter form, so we obtain:
x = 1 + t, y = 2t, and z = 3 + t.