Find a scalar equation of the plane that contains the given point P=(-3,-3,1)and the given normal vector n = (-1,4,7)
To find a scalar equation of the plane that contains the point P=(-3,-3,1) and has the normal vector n=(-1,4,7), we can use the equation of a plane:
Ax + By + Cz = D,
where A, B, C are the components of the normal vector (n) and D is the dot product of the normal vector and any point on the plane (P).
1. Start by substituting the values of the normal vector into the equation:
(-1)x + (4)y + (7)z = D.
2. Next, substitute the coordinates of the given point P into the equation:
(-1)(-3) + (4)(-3) + (7)(1) = D.
3. Simplify the equation:
3 - 12 + 7 = D,
-9 + 7 = D,
-2 = D.
4. Now substitute the value of D back into the original equation to get the scalar equation:
(-1)x + (4)y + (7)z = -2.
Therefore, a scalar equation of the plane that contains the point P=(-3,-3,1) and the normal vector n=(-1,4,7) is -x + 4y + 7z = -2.