A) (15, 11, 4)
B) \[\left( -\frac{17}{3},-\frac{19}{3},1 \right)\]
C) (8, 4, 4)
D) \[\left( -\frac{17}{3},-\frac{19}{3},4 \right)\]
Correct Answer: D
Solution :
The image of a point\[({{x}_{1}},{{y}_{1}},{{z}_{1}})\]in a plane \[ax+by+cz+d=0\]is \[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\] \[=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] Here, image of point (-1, 3, 4) in a plane\[x-2y=0\]is given by \[\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}\] \[=\frac{-2(1\times (-1)+(-2)\times 3+0)}{1+4}\] \[\Rightarrow \]\[\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2(-7)}{5}\] \[\Rightarrow \]\[x=\frac{14}{5}-1=\frac{9}{5},y=-\frac{28}{5}+3=-\frac{13}{5}\] and\[z=4\] Thus, image of point (-1, 3, 4) is\[\left( \frac{9}{5},-\frac{13}{5},4 \right)\] Hence, no option is correct.
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