Let $z=\sqrt{x^2+y^2}$ be cone. Find an equation of each plane tangent to the cone which şs perpendicular to the plane $x+z=5$
I have learnt the solution of such question for parallel at previous question I asked today.
But now, I want to learn properly this questions for perpendicular.
Again, I have its answer. But I cannot comprehend exactly.
The answer is that
Please teach me this again properly. Thank you so much:)
Again I define a function $f(x,y,z)=(x^2/z)+(y^2/z)-z$
And to get a normal vector to the cone at a point (x,y,z) $\nabla f(x,y,z)= (2x/z)ı+(2y/z)j -k $
And then ?
Best Answer
Hint: You can let $f(x,y,z)=\sqrt{x^2+y^2}-z$. Then a normal vector to the tangent plane is given by $\nabla f =\frac{x}{\sqrt{x^2+y^2}}i+\frac{y}{\sqrt{x^2+y^2}}j-k$, and you want this to be orthogonal to the vector $i+k$, which is a normal vector to $x+z=4$ (since the planes are perpendicular). Now set the dot product of these two vectors equal to zero.