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EAMCET Engineering 2009 Solved Paper

Section: Mathematics

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Question : 141 of 160

Marks: +1, -0

The image of the point

(3,2,1)

2 x-y+3 z=7

$(1,2,3)$
$(2,3,1)$
$(3,2,1)$
$(2,1,3)$

Solution:

We know that image

(x, y, z)

of a point

(x_1, y_1, z_1)

in a plane

a x+b y+c z+d=0

is

\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}

= \frac{-2 (a x_1+b y_1+c z_1+d)}{a^2+b^2+c^2}

Here, point is

(3,2,1)

and plane is

2 x-y+3 z=7

.

\therefore \frac{x-3}{2} = \frac{y-2}{-1} = \frac{z-1}{3}

= \frac{-2[2(3)-(2)+3(1)-7]}{2^2+1^2+3^2}

\Rightarrow \frac{x-3}{2} = \frac{y-2}{-1} = \frac{z-1}{3} = -2(0)

\Rightarrow x=3, y=2, z=1

\;

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