Calculus III Question -> Equation of a Line Perpindicular to a Plane thru a certain point 3D?

Question

Find the equation of the line passing through the point B(1,-3,0) and perpindiculat to the plane x + 2y - z = 10

Show me the correct answer and your work for best answer. Thanks

Answer 1

The coefficients of the equation of a plane is the sirection vector of the perpendicular to the plane.
the format of the vector equation of a line in three space looks a lot like the equation of a line in two space.
P = k d + p(0)

Where P is the points on the line (x,y,z)
k the independent variable.
d is the direction vector of the line, in your case
d = (1,2, -1)
And p(0) can be any point on the line. We will use.
(1, -3, 0)
The equation is;
(x,y,z) = k(1,2, -1) + (1, -3, 0)

Answer 2

Find the equation of the line passing through the point
B(1,-3,0) and perpindiculat to the plane
x + 2y - z = 10.

Since the line is perpendicular to the plane, the normal vector n, of the plane is also the directional vector of the line. The normal vector of the plane can be taken from the coefficients of the three variables.

n = <1, 2, -1>

With the directional vector n, of the line and a point
B(1, -3, 0) on the line we can write the equation of the line.

L(t) = B + tn = <1, -3, 0> + t<1, 2, -1>
where t is a scalar ranging over the real numbers