Question

http://screencast.com/t/ePscM8cE

Is there a smart way to do this problem than opening up and simplifying?

Answer 1

@ganeshie8

Answer 2

@iambatman

Answer 3

(u+v-w) . (u-v) x (v-w)

u . (u-v) x (v-w)

Answer 4

cuz the volume of parallelopiped [v-w, u-v, v-2] = 0

Answer 5

we can reduce it further i guess

Answer 6

* [v-w, u-v, v-w] = 0

Answer 7

(u+v-w) . (u-v) x (v-w)

u . (u-v) x (v-w) How you got that?

Answer 8

dot product is distributive

Answer 9

(u+v-w) . (u-v) x (v-w)

(u) . (u-v) x (v-w) + (v-w) . (u-v) x (v-w)

Answer 10

(v-w) . (u-v) x (v-w) becomes 0 as they are perpendicular right?

Answer 11

yes, another easy way is to think of the scalar product as volume of parallelopiped

Answer 12

volume of parallelopiped = \(z \bullet x \times y \)

if any two vectors are same, then the volume is 0

Answer 13

Yeah right.

Answer 14

So how do we proceed from there?

Answer 15

(u) . (u-v) x (v-w)

Answer 16

repeat the same step two more times..

Answer 17

sticking to above volume thing :

(u+v-w) . (u-v) x (v-w)

u . (u-v) x (v-w)

- (u-v) . (u) x (v-w)

v . (u) x (v-w)

Answer 18

one more time and you're done

Answer 19

u. v x w?

Answer 20

careful about signs

Answer 21

actually you're right ! :)

Answer 22

v . (u) x (v-w)
-u.v x(v-w)
-u.v xv-vxw
=u.v x w

Answer 23

Yeah that was what I was thinking. This is a great process :)

Answer 24

while you're at it, below hold in general for triple products :

\[V = [ABC ]= [BCA] = [CAB]\]

\[-V = [ACB ]= [BAC] = [CBA]\]

Answer 25

they look obvious but they're very helpful when simplifying

Answer 26

Yes I have learned that! That is why I didnt ask when you did that manipulation.

Thanks!

Answer 27

u wlcme :)