# Tangent plane at a point

1. Jan 14, 2014

### JonNash

The equation of a tangent plane at the point (1/sqrt3, -1/sqrt3, 1/sqrt3) for a unit sphere with center at origin.

I'm studying for an entrance and this is in the previous question paper MCQ. I've been trying to solve this by studying similar problems but somehow I think the solution here is simpler and does not need gradients to solve. Some of the threads pointed to the possibility that the surface normal vector is given by (2x,2y,2z), but it leads me to no solution. I've tried another solution where the surface normal (not normalized) is given by

P(ø,θ) = (sinøcosθ, sinøsinθ, cosø)

and the vector product of partial derivatives of the above are used but its making no sense to me. Could you please help me out.

Thanks

2. Jan 14, 2014

### consciousness

Welcome to PF! BTW this thread should probably be moved to the math homework forum.

The solution is very simple. Just know that the radius vector joining the point of contact with the plane and the origin will be perpendicular to the plane. This is your "normal' vector. Find this vector in terms of the three unit vectors. Also find the distance between the plane and the origin.
Now, you have the parameters needed to solve this problem.

Which equation do you want? Vector or Cartesian?

3. Jan 14, 2014

### ehild

Yes, with the centre of the sphere in the origin, the normal of the tangent plane is parallel to the vector r0 pointing to Po (1/√3;-1/√3;1/√3). All vectors lying in the tangent plane are perpendicular to n. So the equation of the plane is simply (r-r0)n=0. In Cartesian coordinates, (x-x0)nx+(y-y0)ny+(z-z0)nz=0, where nx=xo, ny=yo and nz=zo in this problem.

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4. Jan 14, 2014

### JonNash

Thanks, I don't know how to move the thread, I will find out and move it asap.

I found the radius vector as r=(1/√3)i - (1/√3)j + (1/√3)k
and the distance between them is 1 (since it is the radius, I even solved it just to be thorough.) Then I have the perpendicular line and the point through which it passes, so the equation of the plane is given by a(x-x0)+b(y-y0)+c(z-z0)=0 and I need the equation in cartesian. So I solved it and I got x-y+z=√3. But now I have a different problem, I have four options and RHS of each is 1/√3, so did I do something wrong?

Thanks ehild that equation put me on the right path, but now as I mentioned above the solution is not among the given choices, is it a printing mistake or did I do something wrong here?

5. Jan 15, 2014

### ehild

Your solution is correct, it must be a misprint in the given solution.

ehild

6. Jan 15, 2014

### JonNash

Thanks ehild. been racking my brains about it.

7. Jan 15, 2014

### consciousness

Also you cant move the thread, that is for the mods to do. :rofl:

Edit: Just crossed my mind but you get 1/√3 on the RHS if you divide your equation by 3 on both sides.

8. Jan 15, 2014

### JonNash

But that would take me farther away from the choices. Just to be clear, here are the choices present.
a)x+y+z=1/√3
b)x-y+z=1/√3
c)x-y-z=1/√3
d)x+y-z=1/√3

If the b) option on the RHS would have been √3 instead of 1/√3, I would be dancing around happy but the answer which I got gets me close to it but matches none of the choices which in turn is inducing a pandemonium in my head.

9. Jan 15, 2014

### consciousness

Yes the options are incorrect.