Question

The angle between the tangent plane to the sphere $x^{2} + y^{2} + z^{2} = 9$ at the point (2,1,2) and the XY - plane is}

• A. $\cos^{-1}(\frac{2}{3})$
• B. $\cos^{-1}(\frac{3}{\sqrt{5}})$
• C. $\cos^{-1}(\frac{4}{9})$
• D. $\cos^{-1}(\frac{\sqrt{2}}{3})$

Hint:

Normal of a sphere at $(x,y,z)$ is just the vector $(x,y,z)$. Normal of XY-plane is always the Z-axis vector ($0,0,1$).

Answer 1

The Correct Option is A

Step 1: Concept
The angle between two planes is the angle between their normal vectors.

Step 2: Meaning
The normal to the tangent plane at $(x_1, y_1, z_1)$ is the vector $(x_1, y_1, z_1)$. For $(2,1,2)$, the normal is $\vec{n_1} = 2i + j + 2k$. The XY-plane has the equation $z = 0$, so its normal is $\vec{n_2} = k$.

Step 3: Analysis
$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(2i + j + 2k) \cdot k|}{\sqrt{2^2 + 1^2 + 2^2} \sqrt{1^2}}$. This becomes $\frac{2}{\sqrt{9} \cdot 1} = \frac{2}{3}$.

Step 4: Conclusion
Therefore, $\theta = \cos^{-1}(\frac{2}{3})$. Final Answer: (A)