CBSE Class 12 Math 2020 Outside Delhi Set 3 Solved Paper

Question : 11 of 11

Marks: +1, -0

Find the image of the point

(-1,3,4)

in the plane

x-2 y=0 .

Solution:

We know that the image

( {\x}, {\y}, {\z})

{\point} ( {\x}_1, {\y}_1 {\z}_1)

In a plane

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

\;{x-x_1}/{a}=\;{y-y_1}/{b}=\;{z-z_1}/{c}=\;{-2 (a x_1+b y_1+c z_1+d) }/{a^2+b^2+c^2}

Thus, the image of point

(-1,3,4)

in a plane

{\x}-2 {\y}=0

\;{ {\x}+1}/{1}=\;{ {\y}-3}/{-2}=\;{ {\z}-4}/{0}=\;{-2[1(-1)+(-2) 3+0(4)]}/{1+0+4}

\;{x+1}/{1}=\;{y-3}/{-2}=\;{z-4}/{0}=\;{-2(-7)}/{5}

x=\;{14}/{5}-1=\;{9}/{5} ; y=\;{-28}/{5}+3=\;{-13}/{5} & z=4

∴

Image of

(-1,3,4)

(\;{9}/{5}, \;{13}/{5}, 4)

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