UPSC NDA Math Model Paper 9

Parallel vector = the sum of the vectors 2

^

i

+4

^

j

−5

^

k

and b

^

i

+2

^

j

+3

^

k

⇒ Parallel vector = (2

^

i

+4

^

j

−5

^

k

) +(b

^

i

+2

^

j

+3

^

k

)
⇒ Parallel vector = (2+b)

^

i

+6

^

j

−2

^

k

Now, unit vector parallel to the sum of the vectors 2

^

i

+4

^

j

−5

^

k

and b

^

i

+2

^

j

+3

^

k

=

(2+b) ^ i +6 ^ j −2 ^ k

√(2+b)2+62+22

Given, scalar product of the vector

^

i

+

^

j

+

^

k

with the unit vector is unity
So, =

(2+b) ^ i +6 ^ j −2 ^ k

√(2+b)2+62+22

. (

^

i

+

^

j

+

^

k

)=1
⇒

(2+b)+6−2

√(2+b)2+62+22

=1
⇒(2+b)+6−2=√(2+b)2+62+22
Squaring both sides, we get
⇒(b+6)2=(2+b)2+62+22
⇒b2+12b+36=b2+4b+44
⇒8b=8
⇒b=1