Question

The resultant vector of VECTOR P and VECTOR Q is VECTOR R. The resultant becomes 2R when VECTOR P is either doubled or reversed in its direction. Prove that P:Q =ROOT 3 :ROOT 2

Answer 1

Explanation:

Using the formula of vectors -

R^2 = P^2 + Q^2 +2|P||Q|CosΘ ———————-(1)

The given condition is, If Q is doubled, R is doubled means -

(2R)^2 = P^2 + (2Q)^2 +2|P||2Q|CosΘ , on solving gives -

4R^2 = P^2 + 4Q^2 +4|P||Q|CosΘ ————(2)

Another given condition is -

If Q is reversed, R is again doubled, means -

(2R)^2 = P^2 + (-Q)^2 +2|P||-Q|Cos(180-Θ)

IMP Points -

1. Reversing a vector means changing its direction by 180.

2. |-Q| = |Q|

3. Cos(180-Θ) = -CosΘ

On solving, we get -

4R^2 = P^2 + Q^2 -2|P||Q|CosΘ ———-(3)

Subtract Eqn(3) from Eqn(2),

0 = 3Q^2 + 6|P||Q|CosΘ

Assuming Q as non-zero vector,

-3|Q| = 6|P|CosΘ

|Q| = -2|P|CosΘ ———————————(4)

Squaring both sides, Q^2 = 4P^2(CosΘ)^2 —————(5)

Now subtract Eqn(1) from Eqn(2) which gives -

3R^2 = 3Q^2 + 2|P||Q|CosΘ ————-(6)

Substituting the value obtained from Eqn(5) and Eqn(4) in Eqn (6),

3R^2 = 3[4P^2(CosΘ)^2] + 2|P|[-2|P|CosΘ]CosΘ

3R^2 = 12P^2(CosΘ)^2 - 4P^2(CosΘ)^2

3R^2 = 8P^2(CosΘ)^2 —————- (7)

Substituting the values of R^2 and Q^2, obtained from Eqn(4) & Eqn(5)

in Eqn(1) -

[8P^2(CosΘ)^2]3 = P^2 + [4P^2(CosΘ)^2] + 2|P|[-2|P|CosΘ]CosΘ

16P^2(CosΘ)^2 = P^2 + 4P^2(CosΘ)^2 – 4P^2(CosΘ)^2

16P^2(CosΘ)^2 = P^2

Assuming P as non-zero vector,

(CosΘ)^2 = 1/16 ———————(8)

We know from Eqn(5), Q^2 = 4P^2(CosΘ)^2

& from Eqn(7), R^2 = [8P^2(CosΘ)^2]/3

Put value of (CosΘ)^2 obtained from Eqn(8) in both the above

mentioned equations -

Q^2 = 1/4*(P^2)

R^2 = 1/6*(P^2)

Now taking the ratio -

P^2 : Q^2 : R^2 = P^2 : 1/4*(P^2) : 1/6*(P^2)

P^2 : Q^2 : R^2 = 1 : 1/4: 1/6

P^2 : Q^2 : R^2 = 12 : 3 : 2 ….. or we can write -

P^2 : Q^2 : R^2 = 6 : 1.5 : 1

Cheers ..!