# Unit Normal Vector

• Jun 12th 2011, 10:01 AM
imagemania
Unit Normal Vector
For a scalar field $s(x,y,z) = x^{2} - y^{2} - z$
find the normal unit vector to a surface s = 0 at the point (1,1,0)

I know i coudl work out grad for s(x,y,z), but not entirely sure how to go beyond this to get a normal unit vector :(
• Jun 12th 2011, 02:44 PM
mr fantastic
Quote:

Originally Posted by imagemania
For a scalar field $s(x,y,z) = x^{2} - y^{2} - z$
find the normal unit vector to a surface s = 0 at the point (1,1,0)

I know i coudl work out grad for s(x,y,z), but not entirely sure how to go beyond this to get a normal unit vector :(

s = 0 => z = x^2 - y^2.

And you should have been taught how to find the vector normal to a surface of the form z = f(x, y), and you should know how to get a unit vector from a vector.

What have you tried, where are you stuck?
• Jun 13th 2011, 01:09 AM
imagemania
Well i know how to do it for parametrics. My attempt for this would follow something i did at a level:
$z-z_{1} = m_{1}(x-x_{1}) + m_{2}(y-y_{1})$
$m_{1} = \frac{\partial dz}{\partial dx} = 2x$
$m_{2} = \frac{\partial dz}{\partial dy} = -2y$

So i assume the normal is one over these. And can sub in the coordinate points, thus obtaning:

$z - 0 = \frac{1}{2}(x-1) - \frac{1}{2}(y-1)$
$z = \frac{x}{2} - \frac{y}{2}$
Is this correct?
• Jun 13th 2011, 01:54 AM
Plato
Quote:

Originally Posted by imagemania
Well i know how to do it for parametrics. My attempt for this would follow something i did at a level:
$z-z_{1} = m_{1}(x-x_{1}) + m_{2}(y-y_{1})$
$z - 0 = \frac{1}{2}(x-1) - \frac{1}{2}(y-1)$
$z = \frac{x}{2} - \frac{y}{2}$
Is this correct?

No it is not correct.
In fact, you did not begin to answer the actual question.

Can you find $\frac{{\nabla s(1,1,0)}}{{\left\| {\nabla s(1,1,0)} \right\|}}~?$
• Jun 13th 2011, 02:51 AM
imagemania
Yes,
$\nabla s = 2x \hat{i} - 2y \hat{j} - 1\hat{k}$
sub in values, do unit vector:
$\frac{2 \hat{i} - 2 \hat{j} - 1\hat{k}}{\sqrt{9}}$

hence:
$\frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3}\hat{k}$
(or could use -ve)

Im guessing this is the tangent unit vector?

Looking at a past question i think i can use the dot product here to ge the normal i.e.
$(x^{2}-1, y^{2}-1, z-0) . (\frac{2}{3},-\frac{2}{3},-\frac{1}{3}) = 0$
Which after expanding i get a similar answer as above :\ (though 2 instead of a half). The reason i chose this method is because this is what we were taught to use
• Jun 13th 2011, 02:53 AM
Also sprach Zarathustra
Quote:

Originally Posted by Plato
No it is not correct.
In fact, you did not begin to answer the actual question.

Can you find $\frac{{\nabla s(1,1,0)}}{{\left\| {\nabla s(1,1,0)} \right\|}}~?$

$\frac{{\nabla s(1,1,0)}}{{\left\| {\nabla s(1,1,0)} \right\|}}=\frac{{\nabla s(1,1,0)}}{\tex{max}{Ds(1,1,0)}}=\frac{\frac{\part ial s}{\partial x}(1,1,0)i+\frac{\partial s}{\partial y}(1,1,0)j+\frac{\partial s}{\partial z}(1,1,0)k}{\sqrt{[\frac{\partial s}{\partial x}(1,1,0)]^2+[\frac{\partial s}{\partial y}(1,1,0)]^2+[\frac{\partial s}{\partial z}(1,1,0)]^2}}$
• Jun 13th 2011, 03:40 AM
imagemania
I assume your partial derivatives should go x, y, z not x,x ,x as that wouldn't make sense.
Isn't that what i did?
which is where the root 9 came from.

But that itself cannot be the normal unit vector surely. Is that not the tangent unit vector and hence this has to be used to obtain the unit normal vector that is at right angles to this tangent (gradient)
• Jun 13th 2011, 03:51 AM
Plato
Quote:

Originally Posted by imagemania
Isn't that what i did?
which is where the root 9 came from.

But that itself cannot be the normal unit vector surely.

Yes it is the same as what you did. And that is the correct answer.

Why do you say that?
Do you have a textbook or a set of lecture notes?
Yes, it was a level surface. You were asked to find the normal unit vector to a surface s = 0 which is clearly a level surface (the "level" being 0). To find a unit normal vector to a surface f(x,y,z)= constant, at a given point, evaluate $\nabla f$ at the given point and divide by its length.