A B C is a triangle, right angled at A . The resultant of the forces acting along overlineA B, overlineB C with magnitudes (1/A B) and (1/A C) respectively is the force along overlineA D, where D is the foot of the perpendicular from A onto B C. The magnitude of the resultant is

Question

A B C is a triangle, right angled at A . The resultant of the forces acting along overlineA B, overlineB C with magnitudes (1/A B) and (1/A C) respectively is the force along overlineA D, where D is the foot of the perpendicular from A onto B C. The magnitude of the resultant is (A) ((AB)(AC)/AB+AC) (B) (1/AB)+(1/AC) (C) (1/AD) (D) (AB2+AC2/(AB)2(AC)2)

Tardigrade Admin · Accepted Answer

Let |BC|=l ∴ In Δ Lambda ABC, l=√AB2+AC2 and tan θ=(AB/AC) ⇒ sin θ=(AB/l) and cos θ=(AC/l) ∴ Resultant vector (1/AB) hati+(1/AC) hatj=((1/l sin θ) hati+(1/l cos θ) hatj) =k AD Now, AD=AC sin θ=l sin θ cos θ =(AB⋅ AC/l) Magnitude of resultant vector =√(1/l2)((1/sin2 θ)+(1/cos2 θ)) =(l/(AB)(AC))=(1/AD)

$\frac{( A B ) ( A C )}{A B + A C}$

$\frac{1}{A B} + \frac{1}{A C}$

$\frac{1}{A D}$

$\frac{A B ^{2} + A C ^{2}}{( A B ) ^{2} ( A C ) ^{2}}$

Q. ABC is a triangle, right angled at A. The resultant of the forces acting along AB,BC with magnitudes AB1​ and AC1​ respectively is the force along AD, where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is

AB+AC(AB)(AC)​

AB1​+AC1​

AD1​

(AB)2(AC)2AB2+AC2​

$\frac{( A B ) ( A C )}{A B + A C}$

$\frac{1}{A B} + \frac{1}{A C}$

$\frac{1}{A D}$

$\frac{A B ^{2} + A C ^{2}}{( A B ) ^{2} ( A C ) ^{2}}$