Question

The plane \(x + y + z = \sqrt{3}\lambda\) touches the sphere \(x^2 + y^2 + z^2 - 2x - 2y - 2z - 6 = 0\) if:

• A. \( \lambda = \sqrt{3} \pm 3 \)
• B. \( \lambda = \sqrt{3} + 3 - \sqrt{2} \)
• C. \( \lambda = \sqrt{3} \pm \frac{1}{3} \)
• D. \( \lambda = \frac{1}{\sqrt{3}} \pm 3 \)

Hint:

To find the center and radius of a sphere from the general equation \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\), the center is \((-u, -v, -w)\) and the radius is \(R = \sqrt{u^2+v^2+w^2-d}\). This is faster than completing the square. In our case, \(2u=-2 \implies u=-1\), \(2v=-2 \implies v=-1\), \(2w=-2 \implies w=-1\), and \(d=-6\). Center is (1,1,1), Radius is \(\sqrt{(-1)^2+(-1)^2+(-1)^2 - (-6)} = \sqrt{1+1+1+6} = \sqrt{9} = 3\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A plane touches a sphere if and only if the perpendicular distance from the center of the sphere to the plane is equal to the radius of the sphere.
Step 2: Key Formula or Approach:
1. Find the center (C) and radius (R) of the sphere from its equation. 2. The distance (d) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is \(d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\). 3. Set the condition \(d = R\) and solve for the unknown parameter \(\lambda\).
Step 3: Detailed Explanation:
1. Find the center and radius of the sphere:
The equation of the sphere is \(x^2 + y^2 + z^2 - 2x - 2y - 2z - 6 = 0\). We can rewrite this by completing the square: \[ (x^2 - 2x + 1) + (y^2 - 2y + 1) + (z^2 - 2z + 1) - 6 - 1 - 1 - 1 = 0 \] \[ (x-1)^2 + (y-1)^2 + (z-1)^2 = 9 \] The center of the sphere is \(C = (1, 1, 1)\). The radius of the sphere is \(R = \sqrt{9} = 3\).
2. Find the distance from the center to the plane:
The equation of the plane is \(x + y + z - \sqrt{3}\lambda = 0\). The center is \((x_0, y_0, z_0) = (1, 1, 1)\). The distance \(d\) is: \[ d = \frac{|1(1) + 1(1) + 1(1) - \sqrt{3}\lambda|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|3 - \sqrt{3}\lambda|}{\sqrt{3}} \] 3. Set distance equal to radius and solve for \(\lambda\):
\[ d = R \implies \frac{|3 - \sqrt{3}\lambda|}{\sqrt{3}} = 3 \] \[ |3 - \sqrt{3}\lambda| = 3\sqrt{3} \] This gives two possible equations: Case 1: \( 3 - \sqrt{3}\lambda = 3\sqrt{3} \) \[ -\sqrt{3}\lambda = 3\sqrt{3} - 3 \] \[ \lambda = -(3 - \frac{3}{\sqrt{3}}) = -(3 - \sqrt{3}) = \sqrt{3} - 3 \] Case 2: \( 3 - \sqrt{3}\lambda = -3\sqrt{3} \) \[ -\sqrt{3}\lambda = -3\sqrt{3} - 3 \] \[ \lambda = 3 + \frac{3}{\sqrt{3}} = 3 + \sqrt{3} \] So, the two possible values for \(\lambda\) are \( \sqrt{3} + 3 \) and \( \sqrt{3} - 3 \). This can be written as \( \lambda = \sqrt{3} \pm 3 \). This seems to match option (A). Let's recheck the options. Ah, the option is \(\lambda = \sqrt{3} \pm 3\). My values are \(3 \pm \sqrt{3}\). Let's recheck the algebra. \( 3 - \sqrt{3}\lambda = 3\sqrt{3} \implies \sqrt{3}\lambda = 3 - 3\sqrt{3} \implies \lambda = \frac{3}{\sqrt{3}} - 3 = \sqrt{3}-3 \). \( 3 - \sqrt{3}\lambda = -3\sqrt{3} \implies \sqrt{3}\lambda = 3 + 3\sqrt{3} \implies \lambda = \frac{3}{\sqrt{3}} + 3 = \sqrt{3}+3 \). So the solutions are indeed \( \lambda = \sqrt{3} \pm 3 \).
Step 4: Final Answer:
The values of \(\lambda\) are \( \sqrt{3} \pm 3 \).