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Find out the unit normal to the surface xy3z2= 4 at (1,12) |
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Answer» Surface xy3z2 = 4 ∴ Normal vector to given surface is ∇(xy3z2) \(=\hat i\frac{\partial}{\partial\mathrm x}\mathrm xy^3z^2+\hat j \frac{\partial}{\partial y}\mathrm x y^3z^2+\hat k\frac{\partial}{\partial z}(\mathrm xy^3z^2)\) = y3z2\(\hat i\) + 3xy2z2\(\hat j\) + 2xy3z\(\hat k\) Normal vector to given surface at (1, 1, 2) is \(\mathrm{\left[\nabla(xy^3z^2)\right]}_{(1, 1, 2)}\) = 4\(\hat i\) + 12\(\hat j\) + 8\(\hat k\) (By putting x = 1, y = 1, z = 2) So, the vector normal to surface xy3z2 = 4 is 4\(\hat i\) + 12\(\hat j\) + 8\(\hat k\). |
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