Find an equation of the plane. the plane through the point (4, 0, 7) and perpendicular to the line x = 4t, y = 9 − t, t = 2 + 2t

4x -y +2z = 30

Step-by-step explanation:

In the standard-form equation of a plane, the coefficients of the variables are the same as those of the direction vector that is normal to the plane. We are given a normal line to the plane we want, so we just need to find the direction vector of that line to be half done. To complete the equation, we need to find the constant that will make the plane go throught the given point.

The equation of the line can be regrouped to parametric form:

... (x, y, z) = (0, 9, 2) + t·(4, -1, 2)

The coefficient of t is the direction vector of the normal to the plane, hence the coefficients of x, y, z in the equation of the plane. The constant in the equation of the plane is the one that allows the given point to be found on the plane:

... 4x -y +2z = 4(4) -(0) +2(7) = 16 +14 = 30

The plane through (4, 0, 7) and perpendicular to (x, y, z) = (0, 9, 2) +t·(4, -1, 2) has equation ...

... 4x -y +2z = 30

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value