Determine whether the lines l1 and l2 are parallel, skew, or intersecting. if they intersect, find the point of intersection. l1: x=3-2t, y=7+4t, z=-3+8tl2: x=-1-u, y=18+3u, z=7+2u

The lines ared skew

Step-by-step explanation:

First we check if they are parallel. To do this we compare the ratios of the slopes of each component, and this should be the same:

As we can see the ratios are different so they are not parallel lines.

Now we test for intersesction:

we have to equate each component and solve it as a system:

From equation 1: u=2t-4 we replace this in eq. 2

From equation 2: t=1/2 replacing both in equation 3 should give us an identity

From equation 3: we reach 1=3 as this is not true, out hypothesis that the lines intersect is wrong.

Since this lines are not paralel and don't intersect we must conclude that the are skew

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the answer is d

hope this