Find the vector and parametric equations for the line through the point P(0, 0, 5) and orthogonal to the plane −1x+3y−3z=1. Vector Form: r

Note that orthogonal to the plane means perpendicular to the plane.

Step-by-step explanation:

-1x+3y-3z=1 can also be written as -1x+3y-3z=0

The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).

Let us find a point on this  line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively

Therefore, the vector equation is given as:

-1(x-0) + 3(y-0) + -3(z-5) = 0

-x + 3y + (-3z+15) = 0

-x + 3y -3z + 15 = 0

Multiply through by - to get a positive x coordinate to give

x - 3y + 3z - 15 = 0

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2