Final answer:
To find the points on the graph of z = 3x²-4y² at which the vector n = (3, 2, 2) is normal to the tangent plane, we need to find the partial derivatives of z with respect to x and y. Equating the partial derivatives with the components of the normal vector, we can find the points on the graph.
Step-by-step explanation:
To find the points on the graph of z = 3x²-4y² at which the vector n = (3, 2, 2) is normal to the tangent plane, we need to find the partial derivatives of z with respect to x and y. The partial derivative of z with respect to x is: ∂z/∂x = 6x and the partial derivative of z with respect to y is: ∂z/∂y = -8y.
Now, we can use the normal vector n = (3, 2, 2) to find the points on the graph where it is normal to the tangent plane. Equating the partial derivatives with the components of the normal vector, we have:
6x = 3 and -8y = 2.
Simplifying these equations, we get:
x = 0.5 and y = -0.25.
Therefore, the points on the graph of z = 3x²-4y² at which the vector n = (3, 2, 2) is normal to the tangent plane are (0.5, -0.25, z), where z is any real number.