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)
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)