NDA Math Vector Algebra

(B) Given

→

a

=

^

i

− 2

^

j

+ 5

^

k

and

→

b

= 2

^

i

+

^

j

− 3

^

k

Now,

→

b

−

→

a

= (2−1)

^

i

+ (1+2)

^

j

+ (−3−5)

^

k

=

^

i

+ 3

^

j

− 8

^

k

3

→

a

+

→

b

= 3(

^

i

− 2

^

j

+ 5

^

k

) + (2

^

i

+

^

j

−3

^

k

)
= (3+2)

^

i

+ (−6+1 )

^

j

+ (15−3)

^

k

= 5

^

i

− 5

^

j

+ 12

^

k

∴ (

→

b

-

→

a

)∙(3

→

a

+

→

b

) = (

^

i

+3

^

j

− 8

^

k

) ∙ (5

^

i

−5

^

j

+12

^

k

)
= 5 − 15 − 96 = −106