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

raj razi
31 Points
11 years ago

A.B=9

A.B=|A|*|B| COSØ

yours katarnak Suresh
43 Points
11 years ago

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.

souvik sonu roy
34 Points
11 years ago

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

lokesh soni
37 Points
11 years ago

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

BAYANA SAGAR
48 Points
11 years ago

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

FITJEE
43 Points
10 years ago

3-2+6=7

Ishita
13 Points
5 years ago
4-2+2=4 this way etfthnjjfdd hhfgj grdh ghbc will din run the yeah dude teach give you the yeah hung day going have fun of him he go get can go to if that's fine you is go is well ha well if so if so I'd will of do if so
3 years ago
Dear student,

a = i + 2j + 3k
|a| = √14
b = 3i – j + 2k
|b| = √14
Hence, a.b = 3 – 2 + 6 = 7
Let Q be the angle between them.
Hence, cosQ = a.b/|a||b| = 7/(√14 x √14)
= 7/14 = ½
or, cosQ = ½
Hence, Q = 60o

Thanks and regards,
Kushagra