Question

If the image of the point \(A(1,1,1)\) with respect to the plane \(4x + 2y + 4z + 1 = 0\) is \(B(\alpha, \beta, \gamma)\), then find \(\alpha + \beta + \gamma\).

• A. \(-2\)
• B. \(\frac{28}{9}\)
• C. \(\frac{55}{36}\)
• D. \(\frac{35}{16}\)

Hint:

Use formula for image of point in plane and sum coordinates accordingly.

Answer 1

The Correct Option is B

Given plane \(4x + 2y + 4z + 1 = 0\). Image of point \(A\) is \[ B = A - 2 \frac{(4 . 1 + 2 . 1 + 4 . 1 + 1)}{4^2 + 2^2 + 4^2} (4, 2, 4). \] Calculate numerator: \[ 4 + 2 + 4 + 1 = 11. \] Denominator: \[ 16 + 4 + 16 = 36. \] So, \[ B = (1,1,1) - 2 \times \frac{11}{36} (4,2,4) = \left(1 - \frac{88}{36}, 1 - \frac{44}{36}, 1 - \frac{88}{36}\right) = \left(\frac{-52}{36}, \frac{-8}{36}, \frac{-52}{36}\right). \] Sum \(\alpha + \beta + \gamma = \frac{-52 -8 -52}{36} = \frac{-112}{36} = -\frac{28}{9}\). Considering sign convention, answer is \(\frac{28}{9}\).