In a right angle $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}$ and sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are respectively, $5 \mathrm{~cm}, 4 \mathrm{~cm}$ and $3 \mathrm{~cm}$ . If a force $\vec{F}$ has moments 0,9 and 16 in N cm . units respectively about vertices $A, B$ and $C$, then magnitude of $\vec{F}$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Chapter_pending

In a right angle

△ ABC, ∠ A = 9 0^{\circ}

and sides

a, b, c

are respectively,

5 cm, 4 cm

and

3 cm

. If a force

F

has moments 0,9 and 16 in N cm . units respectively about vertices

A, B

and

C

, then magnitude of

F

Answer: (C)

Since the moment about

A

is zero, the line of action of

F

passes through

A

. Place

A

at the origin with the right angle there, so

B

lies at distance

A B = 3

along one axis and

C

at distance

A C = 4

along the perpendicular one. For a line through the origin in a direction fixed by

ϕ

, the perpendicular distances from

B

and

C

are

3 sin ϕ

and

4 cos ϕ

, so the moments give

3 F sin ϕ = 9

and

4 F cos ϕ = 16

, that is

F sin ϕ = 3

and

F cos ϕ = 4

. Squaring and adding,

F^{2} = 9 + 16 = 25

, so

F = 5

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option