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

6, 10

9, -13

12, -21

18, -23

step-by-step explanation:

you just add 6 to the x-coordinate and subtract 10 from the y-coordinate.

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)