Manipal Entrance Test 2015 Math Paper

We know that, the image of the point (x1,y1,z1) in the plane ax+by+cz+d=0 is given by

x−1

2

=

y−3

−1

=

z−4

1

=

−2(2−3+4+3)

4+1+1

=

−2(ax1+by1+cz1+d)

a2+b2+c2

So, the image of the point P(1,3,4) in the plane 2x−y+z+3=0 is given by

x−1

2

=

y−3

−1

=

z−4

1

=

−2(2−3+4+3)

4+1+1

⇒

x−1

2

=

y−3

−1

=

z−4

1

=−2
⇒ x=−3,y=5,z=2
Hence, the required point is (-3,5,2).