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

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

1/2*qr*rs

step-by-step explanation: