Problem about tangent plane to surface

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To find points on the surface defined by x^2 + y^2 + z^2 = 7 where the tangent plane is parallel to 2x + 4y + 6z = 1, the normal vectors of both surfaces must be compared. The normal vector of the surface is derived from its gradient, resulting in n1 = (2x, 2y, 2z), while the plane's normal vector is n2 = (2, 4, 6). The cross product of these vectors must equal zero, leading to a system of equations that initially suggests infinite solutions. However, to find specific points, it is essential to consider the sphere's radius and the direction of the normal vector, ultimately leading to the correct points: (1/sqrt(2), sqrt(2), 3/sqrt(2)) and (-1/sqrt(2), -sqrt(2), -3/sqrt(2)). The solution requires combining the conditions of parallelism with the constraint of the sphere's surface.
supermiedos
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Homework Statement



find the points on the surface x^2 + y^2 + z^2 = 7 where its tangent plane is parallel to 2x + 4y + 6z = 1

Homework Equations


Equation of a tangent plane:

fx(x - x0) + fy(y - y0) + fz(z - z0) = 0, where fx means partial derivative of f respect to x
n1 X n2 = 0

The Attempt at a Solution


Two planes are parallel if the cross product of their normal vectors is zero. The normal vector of the surface is its gradient, that is: n1 = 2x i + 2y j + 2z k and the normal vector of the plane is
n2 = 2 i + 4 j + 6 k.

when I do n1 X n2 and equal it to zero, i get a system of 3 equations:

12y -8 z = 0, -12x + 4z = 0, 8x - 4y = 0, but it has a infinity number of solutions (y = 2x, z = 3x). what am I doing wrong?

the solutions according to the book is: (1/sqrt(2), sqrt(2), 3/sqrt(2) ) and (-1/sqrt(2), -sqrt(2), -3/sqrt(2)

Thanks in advance
 
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The points have to be on the surface, so x,y,z fulfil also the equation x^2 + y^2 + z^2 = 7.

ehild
 
Hi Supermiedos
You get an infinity of answers because you just solved the fact that 2 planes should be // to each others
Now you have to put them at specific points
Look at the first equation, it's a sphere with radius √7
So you know that for whichever plane you can think of, there will always be 2 planes // to some reference plane that happen to be at distance √7 from the origin
The second equation gives you the orthogonal vector of the plane, (2, 4, 6)
it really doesn't matter where the plane is, you just care about the direction
so imagine being at the origin, you have a given 'direction' (normalize this vector, it has norm=√14) and you want to 'hit the sphere' at two points, and those should be the one you are expecting
 
Thank you so much ehild and oli4. I tought I "involved" the sphere by using its gradient, but that was not enough of course. I solved it now and I got the answer. :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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