Answer:
P = (-1,1,1)
Step-by-step explanation:
for the line
L (t) = (1 − 2t, 2t − 1, t)
since the vector that is tangent to L is its derivative L'=dL/dt , then
L'= dL/dt (t) = (−2, 2 , 1)
then the vector PQ (P to Q) is
PQ = (x,y,z) - (1, 2, 3) = ( x-1 , y-2 ,z-3)
since P belongs to L → x= 1-2t . y=2t-1 and z=t ,
PQ = ( x-1 , y-2 ,z-3) = (-2t , 2t-3 , t-3)
to be ortogonal to L , PQ should be ortogonal to L' , then the dot product between PQ and L should be 0
PQ * L' = 0
(-2t , 2t-3 , t-3)* (−2, 2 , 1) =0
4t + (4t-6) + (t-3) = 0
9t - 9= 0
t = 1
then
P = L(t=1) = ( 1-2*1 , 2*1-1 , 1 ) = (-1,1,1)
P = (-1,1,1)
