Find parametric equations of the line perpendicular to the lines l1: x = 1 − t, y = t, z = 5t − 4 and l2: x = t + 1, y = −t, z = t − 4 at their point of intersection. (enter your answers as a comma-separated list of equations.) why is this line parallel to the xy-plane? this line is parallel to the xy-plane because its is constant.

Q : X = 6λ+1,Y=6λ,Z=-4 ; λ∈R

This line is parallel to the X-Y plane because in its ''Z'' component the number that multiplies to the parameter λ is 0.

Step-by-step explanation:

First we are going to find the intersection between l1 and l2.

X=X , Y=Y , Z=Z that is how we find the intersection

1-t=t+1 (1)

t=-t (2)

5t-4=t-4 (3)

From equation (2) t=0 (We must verify this in (1) and (2))

Replacing t=0 in l1 or in l2 we find that the intersection is I=(1,0,-4)

To find the direction of the line Q we need to calculate the vector product of the directions of l1 and l2

The direction of l1 is (-1,1,5) and from l2 is (1,-1,1)

The direction is given by the numbers that multiplies the t variable in the equations

(-1,1,5) X (1,-1,1) = (6,6,0)

Notice that (-1,1,5).(6,6,0) = 0 so Q will be perpendicular to l1 and (1,-1,1).(6,6,0)=0 so Q will be perpendicular to l2

We have got the direction from Q and a point that belongs to Q so we can write the equations

Q:(X,Y,Z) = λ(6,6,0)+(1,0,-4) ; λ∈R this is the vectorial equation of Q

To find the parametric equation we need to do the following

(6λ,6λ,0)+(1,0,-4) = (6λ+1,6λ,-4)=(X,Y,Z)

Q : X = 6λ+1,Y=6λ,Z=-4 ; λ∈R

Q//X-Y plane because of its ''Z'' component direction that its equal to 0.

step-by-step explanation: