Question

If the plane $lx + my + nz = p$ touches the sphere $x^{2} + y^{2} + z^{2} = a^{2}$ then}

• A. $p = a$
• B. $p^{2} = a^{2}(l^{2} + m^{2} + n^{2})$
• C. $l^{2} + m^{2} + n^{2} = a^{2}$
• D. $p = a^{2}$

Hint:

Tangency condition: Distance from center = Radius.

Answer 1

The Correct Option is B

Step 1: Concept
A plane touches a sphere if the perpendicular distance from the center of the sphere to the plane is equal to the radius of the sphere.

Step 2: Meaning
The sphere $x^2 + y^2 + z^2 = a^2$ has its center at $(0,0,0)$ and radius $a$.

Step 3: Analysis
The perpendicular distance from $(0,0,0)$ to the plane $lx + my + nz - p = 0$ is $\frac{|l(0) + m(0) + n(0) - p|}{\sqrt{l^2 + m^2 + n^2}} = \frac{|-p|}{\sqrt{l^2 + m^2 + n^2}}$.

Step 4: Conclusion
Setting this equal to radius $a$: $\frac{p}{\sqrt{l^2 + m^2 + n^2}} = a \Rightarrow p = a\sqrt{l^2 + m^2 + n^2}$. Squaring both sides gives $p^{2} = a^{2}(l^{2} + m^{2} + n^{2})$. Final Answer: (B)