Find an equation of the plane.

The plane through the point (2,0,1) and perpendicular to the line x = 3t, y = 2-t, z = 3+4t

The equation of the plane which contains the point (2, 0, 1) and perpendicular to the line is .

First, we use the dot product between the vector slope and a vector that pass through and parallel to the plane to derive the equation of the plane:

We determine the vector slope associated with the line:

(1)

By direct inspection, we see that vector slope is .

Then, we construct the following vector equation by means of the dot product between the vector slope and a vector passing through and parallel to the plane:

(2)

The equation of the plane which contains the point (2, 0, 1) and perpendicular to the line is .

We kindly invite to check this question on equation of the plane: link

Consider the vector equation for the given line,

which has tangent (direction) vector

The normal vector to the plane is parallel to this tangent vector, so we can use it as the normal vector itself.

Then the plane has equation

the answer is 0.071

good luck! : d

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w