Question

linear algebra: cross product in r^3?

is the following pairs of lines perpendicular?
x= 2+2t
y=-3-3t
z=4+4t

and

x=2+t
y=4-t
z=5-t

The answer is no but I don't know how to solve it. I think you're supposed to use the dot product or direction cosines..

Answer 1

x= 2+2t
y=-3-3t
z=4+4t
A vector parallel to this line is
u={2,-3,4}

Answer 2

x=2+t
y=4-t
z=5-t
A vector parallel to this line is
v={1,-1,-1}

Answer 3

u dot v = 2(1) +(-3)(-1) + 4(-1)= 2 +3 -4= 1
so u and v are not perpendicular

Answer 4

Since the dot product of the two parallel vectors is not zero.

Answer 5

Then how about this one?
x=3-t
y=4+4t
z=2+2t

and

x= 2t
y=3-2t
z=4+2t

I did the dot product and it was -6 but the answer said that it was perpendicular..

Answer 6

Check that you copied the problem right.

Answer 7

yes I copied it right. Do you think the book is wrong?

Answer 8

The first line, look at the x should that be x=3+2t?

Answer 9

No, it says x = 3 - t in the book. I'm assuming it's a mistake