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).
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.
To determine a scalar equation of the plane, we need to find the normal vector of the plane.
The given plane is defined by the equation r = (2, -1, 4) + s(2, 0, 3) + t(-3, 2, -5).
The direction vector of the plane is the vector (2, 0, 3).
To find the normal vector of the plane, we can take the cross product of the direction vector of the plane with any other vector in the plane.
Let's take the cross product of (2, 0, 3) with the vector (-3, 2, -5):
(2, 0, 3) x (-3, 2, -5) = (6, -6, 6)
The cross product gives us a normal vector of the plane, which is (6, -6, 6).
Now, we can use the normal vector and a point on the plane (4, 0, -1) to write a scalar equation of the plane.
The scalar equation of a plane is given by: Ax + By + Cz = D, where (A, B, C) is the normal vector of the plane and (x, y, z) is a point on the plane.
Using the normal vector (6, -6, 6) and the point (4, 0, -1), we have:
6x - 6y + 6z = D
Plugging in the coordinates of the point (4, 0, -1):
6(4) - 6(0) + 6(-1) = D
24 - 6 + (-6) = D
12 = D
Therefore, the 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) is:
6x - 6y + 6z = 12.
To determine a scalar equation of the desired plane, we need to find a normal vector to the plane and then use it to write the equation in the form ax + by + cz = d.
To find the normal vector, we can use the fact that the desired plane is parallel to the given plane. Two planes are parallel if and only if their normal vectors are parallel. Therefore, we can use the normal vector of the given plane as the normal vector for the desired plane.
The given plane is defined by the equation r = (2, -1, 4) + s(2, 0, 3) + t(-3, 2, -5), where r = (x, y, z).
The direction ratios of the normal vector of the given plane are the coefficients of x, y, and z in the equation of the given plane. So, the normal vector is (2, 0, 3) x (-3, 2, -5) = (0* (-5) - 3*2, 3*(-5) - 2*(-3), 2*2 - 0*(-3)) = (-6, -11, 4).
Now we have the normal vector of the desired plane, which is (-6, -11, 4). To write the scalar equation of the plane, we choose a point on the plane, which is given as (4, 0, -1).
Substituting these values into the equation ax + by + cz = d, we have -6(x - 4) - 11y + 4(z + 1) = 0.
Simplifying, we get -6x + 24 - 11y + 4z + 4 = 0, which can be further simplified to -6x - 11y + 4z + 28 = 0.
Therefore, 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) is -6x - 11y + 4z + 28 = 0.