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, we need to use the point-normal form equation of a plane. The equation is given by:
n · (r - P) = 0
where n is the normal vector, P is a point on the plane, r is a point on the plane, and · represents dot product.
In this case, the given point P is (-3, -3, 1) and the given normal vector n is (-1, 4, 7). Substituting these values into the equation, we have:
(-1, 4, 7) · (r - (-3, -3, 1)) = 0
Expanding the dot product and simplifying, we get:
-1(r - (-3)) + 4(r - (-3)) + 7(r - 1) = 0
Simplifying further, we have:
-(r + 3) + 4(r + 3) + 7(r - 1) = 0
Now, we distribute and collect like terms:
-r - 3 + 4r + 12 + 7r - 7 = 0
Combining like terms, we get:
11r + 2 = 0
To obtain the final scalar equation of the plane, we can rearrange the equation to isolate r:
11r = -2
r = -2/11
Now, substituting the value of r back into the equation, we have:
x = -2/11
y = -2/11
z = -2/11
Thus, the scalar equation of the plane that contains the given point P=(-3, -3, 1) and the given normal vector n=(-1, 4, 7) is:
11x + 11y + 11z + 2 = 0