Resultant of and is . If is doubled, is doubled but when is reversed(-Q), is again doubled. Find P:Q:R.
Answers
Solution :
Let, θ be the angle between and
Given that, is resultant
Then, R² = P² + Q² + 2PQcosθ ...(i)
When is doubled, gets doubled. Then,
(2R)² = P² + (2Q)² + 2P (2Q) cosθ
⇒ 4R² = P² + 4Q² + 4PQcosθ ...(ii)
When is reversed, gets doubled. Then
(2R)² = P² + Q² + 2PQ cos(π - θ)
⇒ 4R² = P² + Q² - 2PQcosθ ...(iii)
Adding (i) and (iii), we get
R² + 4R² = P² + Q² + 2PQcosθ + P² + Q² - 2PQcosθ
⇒ 5R² = 2P² + 2Q² ...(iv)
Adding (ii) and {(iii) × 2}, we get
4R² + 8R² = P² + 4Q² + 4PQcosθ + 2P² + 2Q² - 4PQcosθ
⇒ 12R² = 3P² + 6Q²
⇒ 4R² = P² + 2Q² ...(v)
Now, (iv) - (v) gives
R² = P²
⇒ R²/P² = 1/1
⇒ R/P = 1/1
⇒ R : P = 1 : 1
⇒ R : P = √2 : √2
Putting R² = P² in (v), we get
4R² = R² + 2Q²
⇒ 3R² = 2Q²
⇒ R²/Q² = 2/3
⇒ R/Q = √2/√3
⇒ R : Q = √2 : √3
Therefore, P : Q : R = √2 : √3 : √2
Suppose theta be 'A'
Let A be an angle between vector P and vector Q.
Given,
vector R is resultant.
R^2 = P^2 + Q^2 + 2PQ cosA _____(1)
When vector Q is doubled, vector R is also doubled.
(2R)^2 = P^2 + (2Q)^2 + 2P × (2Q)^2 × cosA
=> 4R^2 = P^2 + 4Q^2 + 4PQ cosA ____(2)
Now,
When vector Q is reversed, vector R is also doubled.
(2R)^2 = P^2 + Q^2 - 2PQ cos(π - A)
=> 4R^2 = P^2 + Q^2 - 2PQ cosA ______(3)
By adding equation (1) and (3), we get
R^2 + 4R^2 = P^2 + Q^2 + 2PQ cosA + P^2 + Q^2 - 2PQ cosA
Here, ( + 2PQ cosA) and ( - 2PQ cosA) is canceled.
R^2 + 4R^2 = P^2 + Q^2 + P^2 + Q^2
We get,
=> 5R^2 = 2P^2 + 2Q^2 _______________(4)
Now,
equation (4) and (5) gives,
R^2 = P^2
=> R^2 / P^2 = 1/1
=> R/P = 1/1
=> R:P = 1:1
So, R:P = √2:√2
Putting R^2 = P^2 in equation (5), we get
4R^2 = R^2 + 2Q^2
=> 4R^2 - R^2 = 2Q^2
=> 3R^2 = 2Q^2
=> R^2 / Q^2 = 2/3
=> R / Q = √2/√3
=> R : Q = √2:√3
Therefore, P : Q : R = √2 : √3 : √2
Thus, the ratio of P : Q : R = √2 : √3 : √2.