AP EAMCET · Maths · Three Dimensional Geometry
If the image of the point \(A(1,1,1)\) with respect to the plane \(4 x+2 y+4 z+1=0\) is \(\mathrm{B}(\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma=\)
- A \(-2\)
- B \(-\frac{28}{9}\)
- C \(\frac{55}{36}\)
- D \(\frac{35}{16}\)
Answer & Solution
Correct Answer
(B) \(-\frac{28}{9}\)
Step-by-step Solution
Detailed explanation
\(k = \frac{-2(4(1)+2(1)+4(1)+1)}{4^2+2^2+4^2} = \frac{-2(11)}{16+4+16} = \frac{-22}{36} = -\frac{11}{18}\) \(\alpha+\beta+\gamma = (1+1+1) + (4+2+4)k = 3 + 10(-\frac{11}{18}) = 3 - \frac{110}{18} = 3 - \frac{55}{9} = \frac{27-55}{9} = -\frac{28}{9}\)
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