Shortest distance between a point and line given 1 point and the line on parameter free form.

[Nivelo]

Question: Decide the shortest distance between the point (-1,-5,-1) and the line given by the equation 1+x = z/3, y=-6.
formula: distance=abs(PQxV)/abs(V) where PQ and V is vectors


We got Q=(-1,-5,-1) and i get P from the parameterform.I set t equal to the equation and then solved x,y,z to get it on parameterform so that its written like this:
(x,y,z)=P +t*V where P is a point and V is a vector. I now calculate PQ,V and the crossproduct PQXV and calculate the distance. After calculations and simplifying i get sqrt(19)/sqrt(10) but the system says its the wrong answer. Can anyone of u calculate and tell me what answer u got?


My answer:sqrt(19)/sqrt(10)

 

[pka]

Question: Decide the shortest distance between the point (-1,-5,-1) and the line given by the equation 1+x = z/3, y=-6. formula: distance=abs(PQxV)/abs(V) where PQ and V is vectors
We got Q=(-1,-5,-1) and i get P from the parameterform.I set t equal to the equation and then solved x,y,z to get it on parameterform so that its written like this: (x,y,z)=P +t*V where P is a point and V is a vector. I now calculate PQ,V and the crossproduct PQXV and calculate the distance. After calculations and simplifying i get sqrt(19)/sqrt(10) but the system says its the wrong answer. Can anyone of u calculate and tell me what answer u got.
My answer:sqrt(19)/sqrt(10)

I got $\displaystyle \frac{\sqrt{12}}{\sqrt{10}}$ Using $\displaystyle Q: (-1,-5,-1)$ and a point on the line $\displaystyle P: (-1,-6,0)$
The $\displaystyle \overrightarrow {PQ} = \left\langle {0,1,-1} \right\rangle$.

 

[Nivelo]

Thanks for the reply. After seeing your result i did the calculations again, got sqrt(11)/sqrt(10) and the system says its correct!

Cheers