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Physics Question on Motion in a plane

Three particles A B & C start from the origin at the same time; A with U velocity 'a' along x - axis, B with a velocity 'b' along y-axis and C with velocity 'c' in XY plane along the line x = y. The magnitude of 'c' so that the three always remain collinear is :

A

a+y2\frac{a+y}{2}

B

ab\sqrt{ab}

C

aba+b\frac{ab}{a+b}

D

2aba+b\frac{\sqrt{2}ab}{a+b}

Answer

2aba+b\frac{\sqrt{2}ab}{a+b}

Explanation

Solution

comparing the slopes b−c20−c2=b−00−a\frac{b - \frac{c}{\sqrt{2}}}{0 - \frac{c}{\sqrt{2}}} = \frac{b - 0}{0 - a} b−c2c2=ba=ab−ac2=bc2\frac{b - \frac{c}{\sqrt{2}}}{ \frac{c}{\sqrt{2}}} = \frac{b }{a} = ab - \frac{ac}{\sqrt{2}} = \frac{bc}{\sqrt{2}} ∴ ab=c(a+b)2\therefore \, ab = \frac{c (a + b)}{\sqrt{2}} c=2aba+bc = \frac{\sqrt{2} ab}{a + b}