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Find the equation of the tangent plane and normal line at the point (1,2, โˆ’1) tothe surface ๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2 = 12

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Solution

To find the equation of the tangent plane and normal line at the point (1,2, โˆ’1) to the surface ๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2 = 12, we can follow these steps:

Step 1: Calculate the partial derivatives of the given surface equation with respect to x, y, and z. โˆ‚/โˆ‚๐‘ฅ (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 2๐‘ฅ โˆ‚/โˆ‚๐‘ฆ (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 4๐‘ฆ โˆ‚/โˆ‚๐‘ง (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 6๐‘ง

Step 2: Substitute the coordinates of the given point (1,2, โˆ’1) into the partial derivatives. โˆ‚/โˆ‚๐‘ฅ (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 2(1) = 2 โˆ‚/โˆ‚๐‘ฆ (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 4(2) = 8 โˆ‚/โˆ‚๐‘ง (๐‘ฅ2 + 2๐‘ฆ2 + 3๐‘ง2) = 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: ๐‘Ž(๐‘ฅ โˆ’ ๐‘ฅ0) + ๐‘(๐‘ฆ โˆ’ ๐‘ฆ0) + ๐‘(๐‘ง โˆ’ ๐‘ง0) = 0 where (๐‘ฅ0, ๐‘ฆ0, ๐‘ง0) is the given point and (๐‘Ž, ๐‘, ๐‘) are the coefficients obtained from the partial derivatives.

Plugging in the values, we have: 2(๐‘ฅ โˆ’ 1) + 8(๐‘ฆ โˆ’ 2) - 6(๐‘ง + 1) = 0 Simplifying the equation gives: 2๐‘ฅ + 8๐‘ฆ - 6๐‘ง - 10 = 0

Therefore, the equation of the tangent plane at the point (1,2, โˆ’1) is 2๐‘ฅ + 8๐‘ฆ - 6๐‘ง - 10 = 0.

Step 4: Find the equation of the normal line. The normal vector to the tangent plane is given by the coefficients of ๐‘ฅ, ๐‘ฆ, and ๐‘ง 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, โˆ’1) can be written as: ๐‘ฅ = ๐‘ฅ0 + ๐‘Ž๐‘ก ๐‘ฆ = ๐‘ฆ0 + ๐‘๐‘ก ๐‘ง = ๐‘ง0 + ๐‘๐‘ก where (๐‘ฅ0, ๐‘ฆ0, ๐‘ง0) is the given point and (๐‘Ž, ๐‘, ๐‘) are the components of the normal vector.

Plugging in the values, we have: ๐‘ฅ = 1 + 2๐‘ก ๐‘ฆ = 2 + 8๐‘ก ๐‘ง = -1 - 6๐‘ก

Therefore, the equation of the normal line at the point (1,2, โˆ’1) is ๐‘ฅ = 1 + 2๐‘ก, ๐‘ฆ = 2 + 8๐‘ก, ๐‘ง = -1 - 6๐‘ก.

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