Question

Find an equation of the plane.

The plane through the point (1, -1, - 1 ) and parallel to the plane 5x - y - z = 6

Answer 1

The equation of the plane that is parallel to the given plane 5x - y - z = 6 and passes through the point (1, -1, -1) is 5x - y - z = 3. We found this by using the normal vector of the given plane and the coordinates of the point it passes through.

Step-by-step explanation:

To find an equation of the plane that is parallel to the given plane 5x - y - z = 6 and passes through the point (1, -1, -1), we will use the fact that parallel planes have the same normal vector. In this case, the normal vector of both planes is given by the coefficients of x, y, and z in the original plane's equation, which are (5, -1, -1).

Since the desired plane should pass through the point (1, -1, -1), we plug these coordinates into the general equation of a plane Ax + By + Cz = D, where (A, B, C) is the normal vector, to find the value of D. The equation becomes:

5(1) - (-1)(-1) - (-1)(-1) = D

This simplifies to:

5 - 1 - 1 = D

Which yields:

D = 3

So, the equation of the desired plane is:

5x - y - z = 3

Answer 2

Answer: equation of the plane is 5x - y - z - 7 = 0

same as ( 5x - y - z = 7 )

Step-by-step explanation:

Given data;

5x - y - z = 6

therefore vector of plane is n° = ( 5, -1, -1)

so equation will be

n°(x-x₀, y-y₀, z-z₀) = 0

( 5, -1, -1) (x-x₀, y-y₀, z-z₀) = 0

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

5x - 5 - y - 1 - z - 1 = 0

5x - y - z - 7 = 0

Therefore equation of the plane is 5x - y - z - 7 = 0

same as ( 5x - y - z = 7 )