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)

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )