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The mirror image of the point (1,2,3) in a plane is \left ( \frac{-7}{3},-\frac{4}{3},-\frac{1}{3} \right ). which of the following points lies on this plane ?
Option: 1 (1,-1,1)
Option: 2 (-1,-1,1)
Option: 3 (1,1,1)
Option: 4 (-1,-1,-1)
 

Answers (1)

best_answer

 

 

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.

-

 

 

 

P(1,2,3) and Q\left(\frac{-7}{3}, \frac{-4}{3}, \frac{-1}{3}\right)

Direction ratio of normal to the plane \left(1-\frac{-7}{3}, 2-\frac{-4}{3}, 3-\frac{-1}{3}\right)

we get \frac{10}{3}, \frac{10}{3}, \frac{10}{3}

and Midpoint of PQ is \left(\frac{-2}{3}, \frac{1}{3}, \frac{4}{3}\right)

equation of plane x + y + z = 1

Correct Option (1)

Posted by

vishal kumar

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