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Let P be a plane passing through the points (2,1,0), (4,1,1) and (5,0,1)and R be any point (2,1,6).Then the image of R in the plane P is :
Option: 1 (6,5,2)
Option: 2 (6,5,-2)
Option: 3 (4,3,2)
Option: 4 (3,4,-2)
 

Answers (1)

best_answer

 

 

Equation of a plane passing through three non collinear point -

 

Cartesian Form

Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of points A, B and C respectively. 

Let P(x, y z) be any point on the plane.

Then, the vectors \overrightarrow{PA},\;\overrightarrow{BA} and \overrightarrow{CA} are coplanar.
\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left [ \overrightarrow{\mathbf{PA}}\;\;\overrightarrow{\mathbf{BA}}\;\;\overrightarrow{\mathbf{CA}} \right ]=0\\\\\mathrm{\;\;\;}{\begin{vmatrix} \mathbf{x-x_1 }&\mathbf{y-y_1} &\mathbf{z-z_1} \\ \mathbf{x_2-x_1} &\mathbf{y_2-y_1} &\mathbf{z_2-z_1} \\ \mathbf{x_3-x_1} &\mathbf{y_3-y_1} &\mathbf{z_3-z_1} \end{vmatrix}=0}\\\\\text{which is required equation of the plane .}

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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}}} 

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Points of plane P ( 2,1,0), (4,1,1)  and ( 5,0,1)  and point R ( 2,1,6) . 

Then the image of R in the plane P is: 

\begin{vmatrix} x-2 &y-1 &z \\ 2&0 &1 \\ 3& -1& 1 \end{vmatrix} = (x-2)[0+1]-(y-1)[2-3]+ z[-2] \\ \\ \Rightarrow x-2+y-1-2z=0\\ \text{plane} , x+y-2z=3

\\\Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2(2+1-12-3)}{6} \\\Rightarrow(x, y, z)=(6,5,-2)

Correct option (2)

Posted by

Kuldeep Maurya

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