(vector sign-!) (dot product- o) If |!p| = 2 & |!q| = 3 and |3!p - 2!q| = 5, evaluate (!p + 3!q) o (2!p - 5!q)

Question

It is a dot product question. !p ( p1, p2, p3) those kind....

Answer 1

Did you get what modulo_fu... is saying?

Since !p o !p = 4 and !q o !q = 9, when you do the binomial expansion (3!p-2!q)o(3!p-2!q) you get
72-12 !p o !q.
Since the modulus of this vector is 5, this product is 25,
and so
12 !p o !q = 47,
from which you can work out the value of !p o !q.

Hence
(!p + 3!q) o (2!p - 5!q)
=2|p|^2 -15|q|^2 + !p o !q
= -(123 and 1/12)

Can the answer really be so awkward, or did I make a mistake?

And, re the above controversy, I'm sure sahsjing is right.

Answer 2

sahsing is wrong. There's nothing that indicates that the vectors are coplanar or collinear.

Dot products are linear: !a o (!b+!c) = !a o !b + !a o !c
same for your binomial , just expand it.

Also, if the |3p - 2q| = 5 , use pythagorean to get sum of coordinate differences squared.

+ added in response to email request.
hy made an error in copying the problem, but his approach is what I was talking about.

I'm going to change notation to a more common, convenient form
Let
p,q be vectors
p (dot) q = , inner product used in hight math

What you may have missed is that for any vector:
= |v|^2

So, expand that last dot product:
= 2 + - 15
(where I've used the communtivity of dot products to simplify the middle terms)
Now, = |p|^2, which you know
similarly , you know

Oh, I see, we need now. sahsing could be right, I'll need to look at the |3p-2q| = 5.

++ added
Ok, use that equation to get for substitution into what whas derived earlier:
<3p-2q,3p-2q> = 25
9 -12 +4 = 25
= (9*2+4*3-25)/12 = (18+12-25)/12 = -5/12

Now, put this value of back into the derived equation.

Answer 3

enable 'a' and 'd' are integers. Write a² in following way: a² = a² + d² - d² = (a² - d²) + d² = (a+d)(a-d) + d² so we've a² = (a+d)(a-d) + d² .......... (a million) enable a=n+2 and d=a million (n+2)² = (n+2+a million)(n+2-a million) + a million² = a million+(n+3)(n+a million) take sq. root on the two components n+2 = ?(a million+(n+3)(n+a million)) multiply with the help of n n(n+2) = n?(a million+(n+3)(n+a million)) ......... (2) enable f(n)=n(n+2) then f(n+a million)=(n+a million)((n+a million)+2) =(n+a million)(n+3) and (2) will become f(n)= n?(a million+f(n+a million)) .............. (3) this is variety of recursive relation the place f(n) is expressed in terms of f(n+a million). Following the comparable trend way we are able to particular f(n+a million) in terms of f(n+2), then f(n+2) in terms of f(n+3) and so on. like this f(n)= n?(a million+f(n+a million)) f(n+a million)=(n+a million)?(a million+f(n+2)) f(n+2)=(n+2)?(a million+f(n+3)) f(n+3)=(n+3)?(a million+f(n+4)) . . . . . . . . . . . . . beginning from 2nd equation in this sequence, we are able to plug each of those into the previous one to obtain: f(n)= n?(a million+f(n+a million)) = n?(a million+(n+a million)?(a million+f(n+2))) = n?(a million+(n+a million)?(a million+(n+2)?(a million+f(n+3)))) = n?(a million+ (n+a million)?(a million+ (n+2)?(a million+ (n+3)?(a million+ f(n+4))))) . . . . . . . . . . . . . = n ?(a million+ (n+a million)?(a million+ (n+2)?(a million+ (n+3)?(a million+ (n+4)?(a million+...))))) this is n(n+2) = n ?(a million+ (n+a million)?(a million+ (n+2)?(a million+ (n+3)?(a million+ (n+4)?(a million+...))))) putting n=a million we've ?(a million+ 2?(a million+ 3?(a million+ 4?(a million+ 5?(a million+...))))) = 3 ------------------ added notice: (not quickly touching directly to query) id (a million) is foundation of set of rules for on the spot computation of squares devoid of using paper, pencils or calculators. to illustrate: 32² = (32+2)(32-2)+2² = 30*34 + 4 = 30*30+30*4+4=900+a hundred and twenty+4=1024 or fifty six² = 50*sixty two+6² = 3100+36=3136

Answer 4

From the given conditions, we can calculate the angle between p and q.
cos(p^q) = ±(6^2+6^2-5^2)/(2*6*6)

(p+3q)∙(2p-5q)
= 2p^2-15q^2+|p||q|cos(p^q)
--------
modulo_fu,

The two vectors must be co-planar. Who's wrong?

Answer 5

sahsjing is correct.