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The image of the point (–1, 3, 4) in the 3 plane x-2y=0  is

  • Option 1)

    \left ( -\frac{17}{3},-\frac{19}{3},4 \right )\;

  • Option 2)

    (15,11,4)

  • Option 3)

    \left ( -\frac{17}{3},-\frac{19}{3},1 \right )\; \;

  • Option 4)

    \left ( \frac{9}{5},-\frac{13}{5},4 \right )\; \;

 

Answers (1)

best_answer

As we learnt in 

Image of a point -

Let L' be the image of point P\left ( \alpha ,\beta ,\gamma \right ) in the plane ax+by+cz+d=0

L' will be given by the formula 

\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}= \frac{-2\left ( a\alpha +b\beta +c\gamma +d \right )}{a^{2}+b^{2}+c^{2}}

-

 

 To find image we take

\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2\left ( -1-6 \right )}{5}

x=\frac{9}{5} ;\, \, y=\frac{-13}{5}; \, \, z=4


Option 1)

\left ( -\frac{17}{3},-\frac{19}{3},4 \right )\;

Incorrect Option

 

Option 2)

(15,11,4)

Incorrect Option

 

Option 3)

\left ( -\frac{17}{3},-\frac{19}{3},1 \right )\; \;

Incorrect Option

 

Option 4)

\left ( \frac{9}{5},-\frac{13}{5},4 \right )\; \;

Correct Option

 

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Plabita

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