A) show geometrically that, for any scalar k and any vectors u and v,
k(u-v)=ku-kv
B) illustrate for k>0 that k(u+v)=ku+kv
Note: bolded are vectors (arrows on top)
ok so these two questions are quite similar.. I think, but I am still having trouble on "proving" them. I am not sure if my method is O.K. (no answers in back of book), so can someone please verify or demonstrate the correct method. Thanks in advance
A) ku-kv=ku-kv
is this ok? does it prove the q?
B) well I thought about this as for any constant number k, it has to be greater than 0 because otherwise the resultant will be 0 and does not prove ku+kv
what about this question?
ok, ill try to describe my drawings
u and v are vectors |u|=|v|=1
now k(u-v)
not sure how to draw the k part but I drew the (u-v) first
ill describe as a triangle: base is u, v is hypotenuse, (u-v) is height
B) what does it mean by k>0 that k(u+v)=ku+kv
how are u supposed to draw that? doesnt it mean that
ku+kv=ku+kv? im not sure how i would draw these to prove this eqtn
To prove A)
Draw two vectors, u and v.
Draw the vector u - v.
Draw a vector parallel to u - v, of a different length (k).
This vector is k(u - v).
Then draw the vectors ku and kv.
Draw the vector ku - kv.
It should be clear that the vectors k(u - v) and ku - kv are parallel and of the same length. Therefore they are equal.
The same process is used to prove B).
ok, i think I have part A) thanks,
for part B), what does it mean by k>0..
im not sure how to SHOW k>0 by drawing
like.. does it mean that k must exist, for k>0, or, does it mean that -k (cant go in a diff direction) is not allowed?
so I have:
draw vectors ku and kv
then draw ku+kv vector
k(u+v)
draw vectors u and v, then draw vector (u+v), then beside it (parallel) draw k(u+v) but of diff length.
so now my question is, how do you know about the k>0 does it mean k must exist or does it mean that k cannot be going in a diff direction