Question

Among all the points on the sphere 𝑥²+𝑦²+𝑧²=24, the point (𝛼, 𝛽, 𝛾) is closest to the point (1, 2,−1). Then, the value of 𝛼+𝛽+𝛾 equals

• A. 4
• B. -4
• C. 2
• D. -2

Answer 1

The Correct Option is A

To find the point \((\alpha, \beta, \gamma)\) on the sphere \(x^2 + y^2 + z^2 = 24\) that is closest to the point \((1, 2, -1)\), we use the method of Lagrange multipliers or geometric reasoning.

The equation of the sphere is:

\(x^2 + y^2 + z^2 = 24\) 

The distance \(D\) from any point \((x, y, z)\) on the sphere to the point \((1, 2, -1)\) is given by:

\(D = \sqrt{(x-1)^2 + (y-2)^2 + (z+1)^2}\)

Minimizing \(D\) is equivalent to minimizing the square of the distance (to avoid dealing with the square root), so we minimize the function:

\(f(x, y, z) = (x-1)^2 + (y-2)^2 + (z+1)^2\)

Subject to the constraint:

\(g(x, y, z) = x^2 + y^2 + z^2 - 24 = 0\)

We can take the gradient of both the function and the constraint:

• \(\nabla f = (2(x-1), 2(y-2), 2(z+1))\)
• \(\nabla g = (2x, 2y, 2z)\)

Using Lagrange multipliers, we set \(\nabla f = \lambda \nabla g\):

• \(2(x-1) = 2\lambda x\)
• \(2(y-2) = 2\lambda y\)
• \(2(z+1) = 2\lambda z\)

Simplifying, we get:

• \(x - 1 = \lambda x\)
• \(y - 2 = \lambda y\)
• \(z + 1 = \lambda z\)

Expressing each in terms of \(\lambda\):

• \(x = \frac{1}{1-\lambda}\)
• \(y = \frac{2}{1-\lambda}\)
• \(z = \frac{-1}{1-\lambda}\)

Now substitute these \(x\), \(y\), and \(z\) values back into the constraint \(x^2 + y^2 + z^2 = 24\):

\(\left(\frac{1}{1-\lambda}\right)^2 + \left(\frac{2}{1-\lambda}\right)^2 + \left(\frac{-1}{1-\lambda}\right)^2 = 24\)

Simplify the left side:

\(\frac{1 + 4 + 1}{(1-\lambda)^2} = 24\)

Which reduces to:

\(\frac{6}{(1-\lambda)^2} = 24\)

Solving for \(\lambda\):

\((1-\lambda)^2 = \frac{6}{24} = \frac{1}{4}\)

\(1-\lambda = \frac{1}{2} \quad \text{or} \quad 1-\lambda = -\frac{1}{2}\)

\(\lambda = \frac{1}{2} \quad \text{(as negative is not viable in context of square root for distances)}\)

Substitute \(\lambda = \frac{1}{2}\) back to get \(x\), \(y\), \(z\):

• \(x = 2\)
• \(y = 4\)
• \(z = -2\)

Thus, the point \((\alpha, \beta, \gamma)\) is \((2, 4, -2)\), and \(\alpha + \beta + \gamma = 2 + 4 - 2 = 4\).

Hence, the correct answer is 4.