Consider the surface xyz=30. How do you find the unit normal vector to the surface at (2,5,3)?

Dec 29, 2016

$\hat{\vec{n}} = \frac{1}{19} \left(15 \hat{\vec{i}} + 6 \hat{\vec{j}} + 10 \hat{\vec{k}}\right)$

Explanation:

$x y z = 30$

write as

$f \left(x , y , z\right) = x y z - 30 = 0$

vector normal to $\text{ "f(x)" }$is given by $\nabla f \left(x , y , z\right)$

$\nabla f \left(x , y , z\right) = \left(\hat{\vec{i}} \frac{\partial}{\partial x} + \hat{\vec{j}} \frac{\partial}{\partial y} + \hat{\vec{k}} \frac{\partial}{\partial z}\right) \left(x y z - 30\right)$

$\nabla f \left(x , y , z\right) = y z \hat{\vec{i}} + x z \hat{\vec{j}} + x y \hat{\vec{k}}$

$\nabla f \left(2 , 5 , 3\right) = \left(5 \times 3\right) \hat{\vec{i}} + \left(2 \times 3\right) \hat{\vec{j}} + \left(2 \times 5\right) \hat{\vec{k}}$

$\nabla f \left(2 , 5 , 3\right) = 15 \hat{\vec{i}} + 6 \hat{\vec{j}} + 10 \hat{\vec{k}}$

call this normal vector ""vecn

unit vector in this direction is given by

$\hat{\vec{n}} = \frac{\vec{n}}{|} \vec{n} |$

$| \vec{n} | = \sqrt{{15}^{2} + {6}^{2} + {10}^{2}} = 19$

$\hat{\vec{n}} = \frac{1}{19} \left(15 \hat{\vec{i}} + 6 \hat{\vec{j}} + 10 \hat{\vec{k}}\right)$