# Tangential vector of a surface

1. Mar 9, 2009

### caseyjay

Hi all,

If I have a surface x2+3y2+2z2=9, how am I suppose to find all the unit magnitude vectors that are tangential to the surface at point (-1,0,9)?

So far, I am able to find the normal vector 2xi+6yj+4zk and the equation of the plane at (-1,0,9) will therefore be 2x+6y+4z=6. However, then I am stuck and I do not know how can I find the unit magnitude vectors that are tangential to the surface.

Can someone please help me?

Thanks in advance.

Casey.

2. Mar 9, 2009

### HallsofIvy

Staff Emeritus
You are almost finished. All tangential vectors must lie in that tangent plane, of course. The "position vector" of a point in that plane can be written $\vec{r}= (x+ 1)\vec{i}+ y\vec{j}+ (z-9)\vec{k}$ and must satisfy 2(x+1)+ 6y+ 4(z- 9)= 0
You can write, then, z- 9= (1/2)(x+1)+ (3/2)y so the vector equation becomes $\vec{r(x,y)}= (x+1)\vec{i}+ y\vec{j}+ (1/2)(x+ 3y+ 1)\vec{k}$. Divide that by its length to get a unit vector. Of course, you will have x and y in that because you have a two parameter set of vectors in the tangent plane. You could use the "unit" condition $(x+1)^2+ y^2+ (1/4)(x+ 3y+1)= 1$ to reduce to one parameter.

While wr

3. Mar 9, 2009

### caseyjay

Hi HallsofIvy, thanks for your help. At the end of your message, you wrote "While wr". Is there something you would like to continue to explain to me?

4. Mar 9, 2009

### HallsofIvy

Staff Emeritus
Actually, no. I started to say "while writing that", it had occured to m that it would be better to work with the angle the vector makes as parameter but then I figured I had better think about it more! If you let $x= cos(\theta)$ and $y= sin(\theta)$, we can use z= (1/2)(x+1)+ (3/2)y so write z as a function of $\theta$ but then we have the same problem: taking r in the xy-plane equal to 1 doesn't make this a unit vector!

5. Mar 9, 2009

### caseyjay

Oh I see. Speaking of that, how do you derive that the position vector of a point in that plane can be written as:

$\vec{r}= (x+ 1)\vec{i}+ y\vec{j}+ (z-9)\vec{k}$

Is there a formula or derivation that I am missing here?

6. Mar 9, 2009

### caseyjay

Hi HallsofIvy,

I figure out how you derive $\vec{r}= (x+ 1)\vec{i}+ y\vec{j}+ (z-9)\vec{k}$ as the tangential vectors. Since x, y, and z must satisfy 2(x+1)+ 6y+ 4(z- 9)= 0, two possibilities of the tangent vectors are $\vec{r}= \vec{i}+ 3\vec{j}-5\vec{k}$ at (0,3,4) and $\vec{r}= 3\vec{i}+ 5\vec{j}-9\vec{k}$ at (2,5,0).

Since if I substitute (0,3,4) and (2,5,0) separately into 2(x+1)+ 6y+ 4(z- 9), I will obtain 0, hence I presume I am right so far. However, I also remember that the scalar product of the normal vector and the tangent vector should be equal to zero since the angle between them is 90 degrees. But if I perform

$(\vec{i}+ 3\vec{j}-5\vec{k})$*$(-2\vec{i}+ 0\vec{j}+36\vec{k})$

and

$(3\vec{i}+ 5\vec{j}-9\vec{k})$*$(-2\vec{i}+ 0\vec{j}+36\vec{k})$

I won't get 0 as a result.

Why is it so?

NB: * is dot product.

7. Mar 10, 2009

### caseyjay

Hi all,

I think I have solved the problem and thanks for all your help. I have made some fundamental mistakes earlier and hence I am unable to solve it correctly. The solution is as follows:

Surface: x2+3y2+2z2=9

Let f(x,y,z)=x2+3y2+2z2

Therefore $$\nabla$$f(x,y,z)=2x$$\vec{i}$$+6y$$\vec{j}$$+4z$$\vec{k}$$.

At (-1,0,2), f(x,y,z)=-2$$\vec{i}$$+8$$\vec{k}$$. It follows that the equation of the tangent plane is -2(x+1)+8(z-2)=0 and therefore -2x+8z=18.

The tangential vectors on the tangent plane is given by [since it crosses (-1,0,2)]:
(x+1)$$\vec{i}$$+y$$\vec{j}$$+(z-2)$$\vec{k}$$

The unit tangential vector is by dividing the tangential vector with its magnitude.

Hope the above helps and once again thank you to all who have helped in solving this.

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