Find the equation of the tangent plane and normal line at the point (1,2, −1) tothe surface 𝑥2 + 2𝑦2 + 3𝑧2 = 12

To find the equation of the tangent plane and normal line at the point (1,2, −1) to the surface 𝑥2 + 2𝑦2 + 3𝑧2 = 12, we can follow these steps:

Step 1: Calculate the partial derivatives of the given surface equation with respect to x, y, and z. ∂/∂𝑥 (𝑥2 + 2𝑦2 + 3𝑧2) = 2𝑥 ∂/∂𝑦 (𝑥2 + 2𝑦2 + 3𝑧2) = 4𝑦 ∂/∂𝑧 (𝑥2 + 2𝑦2 + 3𝑧2) = 6𝑧

Step 2: Substitute the coordinates of the given point (1,2, −1) into the partial derivatives. ∂/∂𝑥 (𝑥2 + 2𝑦2 + 3𝑧2) = 2(1) = 2 ∂/∂𝑦 (𝑥2 + 2𝑦2 + 3𝑧2) = 4(2) = 8 ∂/∂𝑧 (𝑥2 + 2𝑦2 + 3𝑧2) = 6(-1) = -6

Step 3: Use the values obtained in Step 2 to find the equation of the tangent plane. The equation of a plane can be written as: 𝑎(𝑥 − 𝑥0) + 𝑏(𝑦 − 𝑦0) + 𝑐(𝑧 − 𝑧0) = 0 where (𝑥0, 𝑦0, 𝑧0) is the given point and (𝑎, 𝑏, 𝑐) are the coefficients obtained from the partial derivatives.

Plugging in the values, we have: 2(𝑥 − 1) + 8(𝑦 − 2) - 6(𝑧 + 1) = 0 Simplifying the equation gives: 2𝑥 + 8𝑦 - 6𝑧 - 10 = 0

Therefore, the equation of the tangent plane at the point (1,2, −1) is 2𝑥 + 8𝑦 - 6𝑧 - 10 = 0.

Step 4: Find the equation of the normal line. The normal vector to the tangent plane is given by the coefficients of 𝑥, 𝑦, and 𝑧 in the equation of the tangent plane. Therefore, the normal vector is (2, 8, -6).

The equation of the normal line passing through the point (1,2, −1) can be written as: 𝑥 = 𝑥0 + 𝑎𝑡 𝑦 = 𝑦0 + 𝑏𝑡 𝑧 = 𝑧0 + 𝑐𝑡 where (𝑥0, 𝑦0, 𝑧0) is the given point and (𝑎, 𝑏, 𝑐) are the components of the normal vector.

Plugging in the values, we have: 𝑥 = 1 + 2𝑡 𝑦 = 2 + 8𝑡 𝑧 = -1 - 6𝑡

Therefore, the equation of the normal line at the point (1,2, −1) is 𝑥 = 1 + 2𝑡, 𝑦 = 2 + 8𝑡, 𝑧 = -1 - 6𝑡.