In fig. 28.7 if the coordinates of point `P\ a r e\ (a , b , c)` then Write the coordinates of points A, B, C, D, E and F. Write the coordinates of the feet of the perpendiculars from the point P to the coordinate axes. Write the coordinates of the feet of the perpendicular from the point `P` on the coordinate planes `X Y ,\ Y Z\ a n d\ Z Xdot` Find the perpendicular distances of point `P` from `X Y , Y Z\ a n d\ Z X-` planes. Find the perpendicular distances of the point `P` fro the coordinate axes. Find the coordinates of the reflection of `P` are `(a , b , c)` . Therefore `O A=a ,\ O B=b\ n d\ O C=cdot`

Options

A. `sqrt(a^(2)+b^(2)),sqrt(b^(2)+c^(2)),sqrt(c^(2)+a^(2))`
B. `sqrt(b^(2)+c^(2)),sqrt(c^(2)+a^(2)),sqrt(a^(2)+b^(2))`
C. `sqrt(c^(2)+a^(2)),sqrt(a^(2)+b^(2)),sqrt(b^(2)+c^(2))`
D. `a,b,c`

Answer

B

Explanation

`PA,PB` and `PC` are the perpendicular distances of point `P` from `OX,OY` and `OZ` respectively.
In right angled triangle ADP, we have
`AP^(2)=AD^(2)+DP^(2)`
`impliesPA=sqrt(AD^(2)+DP^(2))`
`impliesPA=sqrt(b^(2)+c^(2)) [ :' AD=OB=b` and `PD=OC=c]`
In right angled triangle PDB right angled at D, we have
`PB^(2)=BD^(2)+PD^(2)`
`impliesPB=sqrt(BD^(2)+PD^(2))`
`=PB=sqrt(a^(2)+c^(2)) [ :' BD=OA=a` and `PD=OC=c]`
In right angled triangle PCF right angled at `F`, we have
`PC^(2)=PF^(2)+CF^(2)`
`impliesPC=sqrt(PF^(2)+CF^(2))`
`impliesPC=sqrt(b^(2)+a^(2)) [ :' PF=AD=OB=b` and `CF=OA=a]`
`impliesPC=sqrt(a^(2)+b^(2))`