find the vector form of the equation of the line in lR^3 that passes through P=(-1,0,3) and is perpendicular to the plane with a general equation x-3y+2z = 5
To find the vector form of the equation of the line in ℝ^3 that passes through the point P=(-1, 0, 3) and is perpendicular to the plane with a general equation x - 3y + 2z = 5, we need to determine a vector that is perpendicular to the plane.
The plane with the general equation x - 3y + 2z = 5 can be rewritten in vector form as:
a⋅(r - r₀) = 0,
where r = <x, y, z> is a vector in the plane, r₀ = <x₀, y₀, z₀> is a point in the plane, and a = <1, -3, 2> is the normal vector to the plane.
So, if we want a line that is perpendicular to this plane, the direction vector of the line should be parallel to the normal vector of the plane.
Therefore, the vector form of the equation of the line can be written as:
r = r₀ + t * a,
where r₀ is a point on the line, a is the direction vector of the line, and t is a scalar.
In this case, r₀ = P = (-1, 0, 3), and a = <1, -3, 2>.
So, the vector form of the equation of the line passing through P=(-1, 0, 3) and perpendicular to the plane x - 3y + 2z = 5 is:
r = (-1, 0, 3) + t * <1, -3, 2>,
where t is a scalar parameter that can take any real value.