Determine a scalar equation of the plane through the point (4, 0 -1) and parallel to the plane r = (2, -1 4) + s(2, 0, 3) + t(-3, 2, -5).

Answer 1

To determine a scalar equation of the plane, we need to find a normal vector to the plane.

The given plane is parallel to the plane defined by the vector equation r = (2, -1, 4) + s(2, 0, 3) + t(-3, 2, -5).

A normal vector to this plane can be found by taking the cross product of the direction vectors of the two lines in the vector equation, (2, 0, 3) and (-3, 2, -5).

The cross product is given by:

(2, 0, 3) × (-3, 2, -5) = (2*-5 - 0*2, 3*(-3) - 2*(-5), 2*2 - 0*(-3)) = (-10, -9, 4).

So the vector (-10, -9, 4) is normal to the plane.

Now, let's find the equation of the plane with this normal vector and passing through the point (4, 0, -1).

A scalar equation of a plane is given by:

Ax + By + Cz = D,

where (A, B, C) is the normal vector to the plane, and (x, y, z) are the coordinates of any point on the plane.

We have (A, B, C) = (-10, -9, 4) and (x, y, z) = (4, 0, -1).

Substituting these values into the scalar equation, we get:

-10x - 9y + 4z = D.

To find the value of D, we substitute the coordinates of the given point (4, 0, -1) into the scalar equation:

-10(4) - 9(0) + 4(-1) = D.

-40 - 4 = D.

D = -44.

Therefore, a scalar equation of the plane passing through the point (4, 0, -1) and parallel to the plane r = (2, -1, 4) + s(2, 0, 3) + t(-3, 2, -5) is:

-10x - 9y + 4z = -44.