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Image of a Point in the Plane - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Image of a Point in the Plane is considered one of the most asked concept.

  • 20 Questions around this concept.

Solve by difficulty

QuestionMEDIUM

 The image of the line\frac{x-1}{3}= \frac{y-3}{1}= \frac{z-4}{-5} in the plane 2x -y+z +3 =0 is the line :

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Let P be the plane passing through the line \frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7} and the point (2,4,-3) 

If the image of the  then  point (-1, 3, 4) in the plane P is (\alpha ,\beta ,\gamma ) then \alpha +\beta +\gamma is equal to

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let (\alpha ,\beta ,\gamma ) be the image of the point P (2, 3, 5) in the plane2x+y-3z=6 then \alpha +\beta +\gamma is equal to

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Concepts Covered - 1

Image of a Point in the Plane

The image of the point P(x1, y1, z1) in the plane ax + by + cz + d = 0 is Q(x3, y3, z3) then coordinates of point Q is given by

\mathbf{\frac{x_3-x_1}{a}=\frac{y_3-y_1}{b}=\frac{z_3-z_1}{c}=-\frac{2\left ( ax_1+by_1+cz_1+d \right )}{{a^2+b^2+c^2}}}

Proof:

Let point Q(x3, y3, z3) is the image of the point P(x1, y1, z1) in the plane π : ax + by + cz + d = 0

Let line PQ meets the plane ax+by+cz+d=0 at point R, direction ratio of normal to plane π are (a, b, c), since, PQ perpendicular to plane π.

So direction ratio of PQ are a, b, c 

\\\Rightarrow \text{equation of line PQ is}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=r\;\;\;\;\text{(say)}\\\\\text { any point on line } P Q \text { may be taken as }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left ( ar+x_1,br+y_1,cr+z_1 \right )\\\mathrm{Let\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}Q\equiv\left ( ar+x_1,br+y_1,cr+z_1 \right )\\\text {since, } R \text { is the middle point of } P Q\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( \frac{x_1+ar+x_1}{2},\frac{y_1+br+y_1}{2},\frac{z_1+cr+z_1 }{2}\right )\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( x_1+\frac{ar}{2},y_1+\frac{br}{2},z_1+\frac{cr }{2}\right )

Since, point R lies in the plane π, we get

\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}R\equiv\left ( x_1+\frac{ar}{2},y_1+\frac{br}{2},z_1+\frac{cr }{2}\right )\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}a\left ( x_1+\frac{ar}{2}\right )+b\left ( y_1+\frac{br}{2} \right )+ c\left ( z_1+\frac{cr }{2} \right )+d=0\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( a^2+b^2+c^2 \right )\frac{r}{2}=-\left ( ax_1+by_1+cz_1+d \right )\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{r}=\frac{-2\left ( ax_1+by_1+cz_1+d \right )}{a^2+b^2+c^2}

In the similar method we can also find the coordinates of point R.

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Image of a Point in the Plane

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Image of a Point in the Plane

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 4.45

Line : 6

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