Discussion Forum

The distance of the point (1,-5,9) from the the plane x-y+z=5 measured along  the line x=y=z is 

Answer:

Explanation:

Equation of line passing through the point (1,-5,9) and parrallel to x=y=z is 

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$  (say)

Thus, any point on this  line is  of the  form $(\lambda+1,\lambda-5,\lambda+9)$

 Now, if $P(\lambda+1,\lambda-5,\lambda+9)$ is the point  of intersection of line and plane, then

 $(\lambda+1)-(\lambda-5)+\lambda+9=5$

$\Rightarrow \lambda+15=5$

   $\Rightarrow \lambda =-10$

$\therefore$  Cordinates of point P are

 (-9,-15,-1)

Hence, required distance

$=\sqrt{(1+9)^{2}+(-5+15)^{2}+(9+1)^{2}}$

$=\sqrt{(10)^{2}+(10)^{2}+(10)^{2}}=10\sqrt{3}$