Question

The image of the point $\left( {5,2,6} \right)$ for the plane $x + y + z = 9$ is
A. $\left( {3, - 5,2} \right)$
B. $\left( {\dfrac{7}{2}, - 1,5} \right)$
C. $\left( {\dfrac{7}{3}, - \dfrac{2}{3},\dfrac{{10}}{3}} \right)$
D. $\left( {\dfrac{7}{3}, - \dfrac{2}{3}, - \dfrac{5}{3}} \right)$

Answer 1

Hint : 1.Mirror image of a point object is at the same distance behind the mirror as object in front of the mirror or we can say that the plane of the point lies at the midpoint of the normal joining image and object. Therefore, we should approach the question taking plane as a mirror point given as an object and line of three-dimensional geometry.
2.This type of question demands us to solve it with the help of a bit of construction for understanding what is required.
3.Draw the demands and try to understand what is given data and what is required.

Complete step-by-step answer :
Let us understand the demands of the question.
We will assemble the given data as follows,
Consider point $P = (x,y,z)$ as image of point $Q = (5,2,6)$ through-plane

\left( {virtual{\text{ }}image{\text{ }}in{\text{ }}theplane} \right) \\\ \\\ $$ whose equation is $ x + y + z = 9 $ Since $ PQ $ poeses through point $ Q $ ,so it has the direction ratio same as normal of the plane $ i.e\,\,\,\,\,\,\,(1,1,1) $ $ [\therefore \,\,\,direction\,\,ratio\,\,of\,\,normal\,\,to\,\,plane\,\,ax + by + cz = d \\\ in\,(a,b,c)\,\,\,so\,\,x + y + z = 9\,\,in\,\,(1,1,1)] \\\ $ So, the equation of the line $ PQ $ passing through $ (5,2,6) $ having direction ratio $ (1,1,1) $ Is $$\dfrac{{x - 5}}{1} = \dfrac{{y - 2}}{1} = \dfrac{{z - 6}}{1} = \lambda $$ Since equation of line in $ 3 - d $ , having direction ration $ (a,b,c) $ and passing through a point $$({x_1},{y_1},{z_1})$$ is $$\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c} = \lambda $$ Let us find the general point on line $ 1\lambda $ $ \dfrac{{x - 5}}{1} = \lambda \\\ x = \lambda + 5 \\\ $ , $ \dfrac{{y - 2}}{1} = \lambda \\\ y = \lambda + 2 \\\ $ , $ \dfrac{{z - 6}}{1} = \lambda \\\ z = \lambda + 6 \\\ $ The point $ PQ $ is $ (x,y,z)\,\,is\,\,(\lambda + 5,\,\,\lambda + 2,\,\,\lambda + 6) $  From the figure, clearly, The midpoint of $ PQ $ is $ \Rightarrow \left( {\dfrac{{x + 5}}{2}\,,\,\dfrac{{y + 2}}{2}\,,\,\dfrac{{z + 6}}{2}} \right) = \left( {\dfrac{{\lambda + 5 + 5}}{2}\,,\,\dfrac{{\lambda + 2 + 2}}{2}\,,\,\dfrac{{\lambda + 6 + 6}}{2}} \right) $ Since midpoint of line joining $ \Rightarrow ({x_1},{y_1})\,\,\& \,\,({x_2},{y_2})\,\,in\,\,\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\,\dfrac{{{z_1} + {z_2}}}{2}} \right) $ $ = \,\,\,\left( {\dfrac{{\lambda + 10}}{2},\dfrac{{\lambda + 4}}{2},\,\dfrac{{\lambda + 12}}{2}} \right) $ Since midpoint $ PQ $ lie on all plane $ x + y + z = 9 $ So, it will satisfy the equation of plane $ (x + y + z = 9) $ Now, we will put point $ \,\,\left( {\dfrac{{\lambda + 10}}{2},\dfrac{{\lambda + 4}}{2},\,\dfrac{{\lambda + 12}}{2}} \right) $ In equation $ x + y + z = 9 $ to find value of $ \lambda $ $ \dfrac{{\lambda + 10}}{2},\dfrac{{\lambda + 4}}{2},\,\dfrac{{\lambda + 12}}{2} = 9 $

\Rightarrow \dfrac{{\lambda + 10 + \lambda 4 + \lambda + 12}}{2} = 9 \\
\Rightarrow \dfrac{{3\lambda + 26}}{2} = 9 \\
\Rightarrow 3\lambda + 26 = 18 \\
\Rightarrow 3\lambda = 18 - 26 \\
\Rightarrow 3\lambda = - 8 \\
\Rightarrow \lambda = \dfrac{{ - 8}}{3} \\

So, the required point $ Q $ can be obtained Here, we will use the value of $ \lambda $ to solve further Putting value of $$\lambda = \dfrac{{ - 8}}{3}$$ in coordinate of Point $ \Rightarrow Q = (\lambda + 5,\lambda + 2,\lambda + 6) \\\ \,\, = \,\,\left( {\dfrac{{ - 8}}{3} + 5,\dfrac{{ - 8}}{3} + 2,\dfrac{{ - 8}}{3} + 6} \right) \\\ \,\,\,\,\,\,\,\left( {\dfrac{{ - 8 + 15}}{3},\dfrac{{ - 8 + 6}}{3},\dfrac{{ - 8 + 18}}{3}} \right) \\\ \,\,\,\,\,\,\,\left( {\dfrac{7}{3},\dfrac{{ - 2}}{3},\dfrac{{10}}{3}} \right) \\\ $ So, the required point is $ \,\left( {\dfrac{7}{3},\dfrac{{ - 2}}{3},\dfrac{{10}}{3}} \right) $ **So, the correct answer is “Option C”.** **Note** : The question related to plane and line on the concept of image should be kept in mind that image is at the equal distance from the mirror as object and while solving them we should look for direction ratio of related lines in three-dimensional geometry. looking In these types of questions, there are chances of calculation mistakes. Students can make mistakes in the construction. Though the naming in construction may differ for the ease of understanding the concept but the answer demanded by the question has been fulfilled on the same line.