Question

Find direction of cosines of u x v for the vectors u and v in the accompanying figure.

Answer 1

experimentX. there need help.

Answer 2

change in coordinates/ change in length

Answer 3

1/sqrt(3),1/sqrt(3),1/sqrt(3) for v

Answer 4

1/sqrt(2),1/sqrt(2),0 for u

Answer 5

sorry
change in coordinates/length

Answer 6

I know how to find if dot product is given. but don't know the formula for cross product.

Answer 7

just cross multiply them

ixi=0 (all same unit vectors have zero cross product)
ixj=k ... follow that cycle
jxi=-k ... if not on that cycle

Answer 8

Yes I know this all.

What do you mean by change in coordinates/change in length?

Answer 9

\[ (x_2 - x_1)/\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (x_2 - z_1)^2},\]
\[ (y_2 - y_1)/\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (x_2 - z_1)^2}, \]
\[(z_2 - z_1)/\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (x_2 - z_1)^2} \]

Answer 10

OK Thanks. I have visualized. :)

\[\large \cos \theta=\frac{u \times v}{||u \times v||}\]

Answer 11

I haven't seen that relation

Answer 12

But this gives answer right.

Answer 13

Oh sorry ... i misread!!
most probably from the beginning of the question

Answer 14

hmmm...

Answer 15

direction cosines are given by the components of
\[ \frac{u \times v}{||u \times v||} \]
i mean ,,, i component will give you l, j component will give you m, and k component will give you k

Answer 16

u = <1,1,0>
v = <1,1,1>
take cross product ... divide it's individual component by it's magnitude, you will have direction cosines.
still equation it to cos theta is wrong

Answer 17

Jeez ... why do i always mistype
still equationing it to cos theta is wrong

Answer 18

But question is asking for direction of cosines. What does it mean? Does it not mean to find cos(theta)?

Answer 19

no ...

Answer 20

OH... I see

I got now.

You mean
\[\text{direction of cosine}=\frac{u \times v}{||u||}\]This is for u.

And similarly for v.

Am I right now?

Answer 21

No ...
u and v are vectors (or lines) ... they have direction cosines ... i have posted them on third and second post

Answer 22

cross product of two vectors is another vector perpendicular to it ... you need to find similar set of numbers.
there's two way you can do it.

Answer 23

I know there are two ways. And understood. I have seen your formula and noted down to me. OK. so these are direction of cosines.

Answer 24

yes of ... V and U
not of VxU

Answer 25

Yes yes.. OK thanx buddy. :)

Answer 26

for direction cosine of VxU,
method one is somewhat like you did

Answer 27

my suggestion .. take a good lesson from youtube