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

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

Grade:12

8 Answers

raj razi
31 Points
8 years ago

A.B=9

A.B=|A|*|B| COSØ

yours katarnak Suresh
43 Points
8 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
8 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
8 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
8 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
8 years ago

ANSWER

3-2+6=7

Ishita
13 Points
3 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
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem.
 
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

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