a=i+2j+3k,b=3i-j+2k find dot product and angle between two vectors?

A.B=9

A.B=|A|*|B| COSØ

Approved

When the two vectors are placed this way, the vectors form the angle q . ... The
scalar product (dot product) A · B (A dot B) of vectors A and B is the ... Given A = i
- j + k and B = 3i + 2j - k, find (a) A · B, (b) | A |, | B |, (c) the cosine of the angle
between A and ... Find the vector projection, Proj A B when A = 5j - 3k and B = i +
j + k.

Approved

a=i+2j+3k

b=3i-j+2k

we know that  i.i=1 , j.j=1 , k.k=1. and i.j=j.i=i.k=k.i=j.k=k.j=0.

a.b=3-2+6=7

a.b=abcosθ

cosθ=(a.b)÷ab

{a=√a.a, b=√b.b}  [ a=√14=b}

cosθ=7÷(14)=.5

θ=60

Approved

a.b = lal*lbl* cosx

and  i.i=k.k=j.j=1

and i.j=i.k=j.k=0

hence a.b=3-2+6=7

and lal*lbl=sqrt(14)*sqrt(14)=14

hence 7=14*cosx

cosx = .5

x=60degree

DOT PRODUCT OF TWO VECTORS:

a=i+2j+3k,b=3i-j+2k then a.b=(1)(3)+(2)(-1)+(3)(2)=3-2+6=7

let  alpha be the angle between the vectiors a and b. then cos(alpha)=a.b/modulus(a).modulus(b)

=7/root14.root14

=7/14=1/2

=cos(alpha)=cos60

=alpha=60

ANSWER

3-2+6=7