UPSC NDA Math Model Paper 5

Let's say that the forces acting on the particle are

→

P

=2

^

i

−5

^

j

+6

^

k

and

→

Q

=−

^

i

+2

^

j

−

^

k

.
∴ The resulting force acting on the particle will be

→

F

=

→

P

+

→

Q

→

F

=(2

^

i

−5

^

j

+6

^

k

)+(−

^

i

+2

^

j

−

^

k

)
Since the particle is moved from the point 4

^

i

−3

^

j

−2

^

k

to the point 6

^

i

+

^

j

−3

^

k

, the displacement vector

→

D

will be:

→

D

=

→

AB

=

→

B

−

→

A

=(6

^

i

+

^

j

−3

^

k

)−(4

^

i

−3

^

j

−2

^

k

)
⇒

→

D

=2

^

i

+4

^

j

−

^

k

.
And finally, the work done W will be:
W=

→

F

.

→

D

=(

^

i

−3

^

j

+5

^

k

) .(2

^

i

+4

^

j

−

^

k

)
⇒W=(1)(2)+(−3)(4)+(5)(−1)
⇒W=2−12−5=−15‌units