Finding the Unique Perpendicular Vector at P=(3,4,5)

[rado5]

Is it correct? unique perpendicular vector

Homework Statement

Find the unique vector which is perpendicular on z= $$\sqrt{x^2+y^2}$$ at P=(3,4,5)

Homework Equations

$$\frac{\nabla F}{\left| \nabla F\left|}$$

The Attempt at a Solution

The answer is $$\frac{\nabla F}{\left| \nabla F\left|}$$ and F(x,y,z)=$$\sqrt{x^2+y^2}$$-z so the answer is $$\frac{\nabla F}{\left| \nabla F\left|}$$= ($$\frac{0.6}{\sqrt{2}}$$ , $$\frac{0.8}{\sqrt{2}}$$ , $$\frac{-1}{\sqrt{2}}$$)

 

[mighty2000]

Hello, thank you for your question. Your solution is correct. The unique perpendicular vector at point P=(3,4,5) is indeed (\frac{0.6}{\sqrt{2}} , \frac{0.8}{\sqrt{2}} , \frac{-1}{\sqrt{2}}). This can be verified by taking the dot product of this vector with the gradient of F(x,y,z) at P, which will give a value of zero, indicating perpendicularity. Additionally, this vector is unique because it is the only vector that is perpendicular to the surface defined by z= \sqrt{x^2+y^2} at point P. Great job on your solution!