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Tangential vector of a surface

  1. Mar 9, 2009 #1
    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.

  2. jcsd
  3. Mar 9, 2009 #2


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    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 [itex]\vec{r}= (x+ 1)\vec{i}+ y\vec{j}+ (z-9)\vec{k}[/itex] 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 [itex]\vec{r(x,y)}= (x+1)\vec{i}+ y\vec{j}+ (1/2)(x+ 3y+ 1)\vec{k}[/itex]. 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 [itex](x+1)^2+ y^2+ (1/4)(x+ 3y+1)= 1[/itex] to reduce to one parameter.

    While wr
  4. Mar 9, 2009 #3
    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?
  5. Mar 9, 2009 #4


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    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 [itex]x= cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex], we can use z= (1/2)(x+1)+ (3/2)y so write z as a function of [itex]\theta[/itex] but then we have the same problem: taking r in the xy-plane equal to 1 doesn't make this a unit vector!
  6. Mar 9, 2009 #5
    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?
  7. Mar 9, 2009 #6
    Hi HallsofIvy,

    I figure out how you derive [itex]\vec{r}= (x+ 1)\vec{i}+ y\vec{j}+ (z-9)\vec{k}[/itex] 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 [itex]\vec{r}= \vec{i}+ 3\vec{j}-5\vec{k}[/itex] at (0,3,4) and [itex]\vec{r}= 3\vec{i}+ 5\vec{j}-9\vec{k}[/itex] 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

    [itex](\vec{i}+ 3\vec{j}-5\vec{k})[/itex]*[itex](-2\vec{i}+ 0\vec{j}+36\vec{k})[/itex]


    [itex](3\vec{i}+ 5\vec{j}-9\vec{k})[/itex]*[itex](-2\vec{i}+ 0\vec{j}+36\vec{k})[/itex]

    I won't get 0 as a result.

    Why is it so?

    NB: * is dot product.
  8. Mar 10, 2009 #7
    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 [tex]\nabla[/tex]f(x,y,z)=2x[tex]\vec{i}[/tex]+6y[tex]\vec{j}[/tex]+4z[tex]\vec{k}[/tex].

    At (-1,0,2), f(x,y,z)=-2[tex]\vec{i}[/tex]+8[tex]\vec{k}[/tex]. 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)]:

    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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