Question

find the distance of the point (2,1,-1) from the plane x-2y +4z =0​

Answer 1

Answer:

x

1

,y

1

,z

1

)=(1,2,−1)

⇒ax+by+cz+d=x−2y+4z−10=0

Required perpendicular distance:

=

a

2

+b

2

+c

2

∣ax

1

+by

1

+cz

1

+d∣

=

1

2

+(−2)

2

+4

2

∣1(1)+(−2)(2)+4(−1)−10∣

=

21

17