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# Understand the problem

In $\Bbb R^3$ the cosine of acute angle between the surfaces $x^2+y^2+z^2-9=0$ and $z-x^2-y^2+3=0$ at the point $(2,1,2)$ is
1. $\frac{8}{5\sqrt{21}}$
2. $\frac{10}{5\sqrt{21}}$
3. $\frac{8}{3\sqrt{21}}$
4. $\frac{10}{3\sqrt{21}}$
##### Source of the problem
IIT JAM 2018 Qn no 6
##### Topic
Multivable calculus
Easy
##### Suggested Book
 Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2 Tom M. Apostol

Do you really need a hint? Try it first!

If we are asked to give the angle between two lines then it is very easy to calculate but our forehead will get skinned whenever we will be asked to find out the acute angle between two lines and even worse if we are asked to find the angle between two surfaces.   Surprisingly it is not very hard if think stepwise. Observe when you are asked to find out the angle between two lines you calculate it in terms slope. So basically you are firing putting the gun on someone else’s shoulder. Here the question is to find that shoulder when it comes in finding the angle between two curves. Observe from the conception of the intersection of two curves that the tangent line of those curves also intersects and we have their corresponding slopes. Bingo! why not calculating the acute angle of the tangent lines and call them the angle between two curves.   Now can you think how to calculate the acute angle between to surfaces?
The acute angle between two surfaces would be the acute angle between their tangent plane. You can stop here and try to do the problem by your own otherwise continue…   The main idea of finding tangent planes revolves around finding gradient of the corresponding surfaces. (For more info see question no 5).   Can you calculate the gradient of the surfaces $x^2+y^2+z^2-9$ and $z-x^2-y^2+3$ at $(2,1,2)$?
The gradient of the surfaces $f=x^2+y^2+z^2-9$ and $g=z-x^2-y^2+3$ at $(2,1,2)$ are $n_1=f_xi +f_yj+f_zk$ and $n_2=g_xi +g_yj+g_zk$ at $(2,1,2)$ which is $n_1=2xi+2yj+2zk=4i+2j+4k$  and $n_2=-2xi-2yj+k=-4i-2j+k$.   Now given these two gradients, can you find out the angle between them?
$n_1=2xi+2yj+2zk=4i+2j+4k$  and $n_2=-2xi-2yj+k=-4i-2j+k$.

This follows the cosine angles between two gradient is $cos \theta=|\frac{n_1.n_2}{|n_1||n_2|}|=|\frac{-16-4+4}{\sqrt{36 \times 21}}|=\frac{8}{3\sqrt{21}}$

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