The line (x, y, z) = (1 − 2t, 2t − 1, t), with t as the parameter, is denoted by l. the point q is (1, 2, 3).

a) find a point p on l such that the line from p to q is orthogonal (perpendicular) to l.

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)

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.

positive

step-by-step explanation:

the quadrant is facing upwards