Normal vector problem

[Elena Baby]

I have a problem:
"Prove that any normal vector of z^2=x^2+y^2 hits the z axis."
(I have translated this problem to English so forgive me if it's not understandable.)



What I did was I found the gradient which is equal to 2zk-2xi-2yj=0.
I set x and y equal to 0,and z therefore is equal to 0.
(I'm new to tangent planes and gradients btw)


Please can someone tell me where is wrong and how can I improve my answer?

 

[Dr.Peterson]

Since a vector has no location, I think a better translation would be:

Prove that any line normal to the surface z^2=x^2+y^2 intersects the z-axis.​

To do this, you need more than the gradient at a point on the surface (and you certainly can't set x and y to zero for that point).

Rather, write an equation for the normal line at an arbitrary point (a,b,c) on the surface.

 

[HallsofIvy]

We can write \(\displaystyle z^2= x^2+ y^2\) as \(\displaystyle x= r cos(t)\), \(\displaystyle y= r sin(t)\), \(\displaystyle z= r\).

Taking the derivative with respect to r, \(\displaystyle \left< cos(t), sin(t), 1\right>\) is a tangent vector to the surface. Taking the derivative with respect to \(\displaystyle \theta\), \(\displaystyle \left<-r sin(t), r cos(t), 0\right>\) is also a tangent vector. The cross product
\(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -r sin(t) & r cos(t) & 0 \\ cos(t) & sin(t) & 1 \end{array}\right|= \left<r cos(t), r sin(t), -r\right>\) is the normal to the surface at the point \(\displaystyle \left(r cos(t), rsin(t),r\right>\).

A line normal to the surface at that point is \(\displaystyle x= r cos(t)s+ rcos(t)= rcos(t)(s+ 1)\), \(\displaystyle y= rsin(t)s+rsin(t)= rsin(t)(s+ 1)\), \(\displaystyle z= -rs+ r= r(1- s)\) where "s" is the parameter.

 

[Elena Baby]

Thank you guys!?

 

[Elena Baby]

Since a vector has no location, I think a better translation would be:

Prove that any line normal to the surface z^2=x^2+y^2 intersects the z-axis.​

To do this, you need more than the gradient at a point on the surface (and you certainly can't set x and y to zero for that point).

Rather, write an equation for the normal line at an arbitrary point (a,b,c) on the surface.

You are right that's line.btw Can I solve it without the actual location?

 

[lev888]

You are right that's line.btw Can I solve it without the actual location?

The problem asks you to prove something, not solve. You can plug in a particular point and show that a line passing through that point intersects the z axis. But you need to prove that this is true for any point. So, given an arbitrary point (a,b,c) write an equation for the line, as suggested above. Then show that this line intersects the z axis.

 

[Elena Baby]

The problem asks you to prove something, not solve. You can plug in a particular point and show that a line passing through that point intersects the z axis. But you need to prove that this is true for any point. So, given an arbitrary point (a,b,c) write an equation for the line, as suggested above. Then show that this line intersects the z axis.

Your explaination was amazing!Thank you!