When two vectors of magnitude p and q are inclined at an angle theta the magnitude of the resultant is to be when inclination is changed to 180 - theta the magnitude of the resultant is half find the ratio of p to q?
Answers
-1/3
as costheta is + & cos180-theta is -ve so the ratio will be -1/3
Answer:
k = (10cosθ ± √(100Cos²θ - 36) )/6
Explanation:
When two vectors of magnitude p and q are inclined at an angle theta the magnitude of the resultant is to be when inclination is changed to 180 - theta the magnitude of the resultant is half find the ratio of p to q?
Let say P is vector along x axis
and vector q is at θ
assuming θ is acute angle
Resultant vector = p + qcosθ & q sinθ
Resultant magnitude = √( (p + qcosθ)² + (q sinθ)²
= √p² + q²Cos²θ + 2pqcosθ + q²Sin²θ
= √p² + q² + 2pqcosθ
Now vector q is at 180 -θ
as cos(180-θ) = -cosθ & Sin(180-θ) = Sinθ
Resultant vector = p - qcosθ & q sinθ
Resultant magnitude = √( (p - qcosθ)² + (q sinθ)²
= √p² + q²Cos²θ - 2pqcosθ + q²Sin²θ
= √p² + q² - 2pqcosθ
2 (√p² + q² - 2pqcosθ) = √p² + q² + 2pqcosθ
Squaring both sides
=> 4 (p² + q² - 2pqcosθ) = p² + q² + 2pqcosθ
=> 4p² + 4q² - 8pqcosθ = p² + q² + 2pqcosθ
=> 10pqcosθ = 3(p² + q²)
=> 3p² + 3q² - 10pqcosθ = 0
Let say p/q = k ( ration of magnitude)
=> p = qk
=>3q²k² + 3q² - 10kq²cosθ = 0
dividing by q²
=> 3k² - 10kcosθ + 3 = 0
=> k = (10cosθ ± √(100Cos²θ - 36) )/6