Question

pls solve this..
vectors class 11
1st correct answer will be marked as Brainliest.​

Answer 1

Answer:

vector=vec

a vec=b Vec -c vec

c vec=b vec-a vec

squaring on both sides,

${c \: vec}^{2} = {(b \: vec - a \: vec)}^{2}$

${c}^{2} = {b}^{2} - 2 \times a \: vec \times b \: vec + {a}^{2}$

$a \: vec \times b \: vec = ( {a}^{2} + {b}^{2} - {c}^{2}) \div 2$

now,

angle between a vec and b vec=

$cosx = a \: vec \times b \: vec \div ab$

putting valu of a vec* b vec, then

$x = {cos}^{ - 1} ( {a}^{2}+ {b}^{2} - {c}^{2} ) \div 2ab$

ans....

hope helps you!!!?

thank you!!!

Answer 2

vector=vec

a vec=b Vec -c vec

c vec=b vec-a vec

squaring on both sides,

{c \: vec}^{2} = {(b \: vec - a \: vec)}^{2}cvec

2

=(bvec−avec)

2

{c}^{2} = {b}^{2} - 2 \times a \: vec \times b \: vec + {a}^{2}c

2

=b

2

−2×avec×bvec+a

2

a \: vec \times b \: vec = ( {a}^{2} + {b}^{2} - {c}^{2}) \div 2avec×bvec=(a

2

+b

2

−c

2

)÷2

now,

angle between a vec and b vec=

cosx = a \: vec \times b \: vec \div abcosx=avec×bvec÷ab

putting valu of a vec* b vec, then

x = {cos}^{ - 1} ( {a}^{2}+ {b}^{2} - {c}^{2} ) \div 2abx=cos

−1

(a

2

+b

2

−c

2

)÷2ab

ans....

hope helps you!!!