Three particles, each of mass m are placed at the vertices of a right angled triangle as shown in figure. The position vector of the center of mass of the system is:
(O is the origin and $^i,^j,^k$ are unit vectors)

\frac{2}{3} (a^i + b^j)

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\frac{2}{3} (a^i - b^j)

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\frac{1}{3} (a^i + b^j)

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\frac{1}{3} (a^i - b^j)

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Solution

The correct option is B $\frac{1}{3} (a^i + b^j)$
Position of centre of mass is given by, ${\to x}_{c m} = \frac{\sum m_{i} {\to x}_{i}}{\sum m_{i}}$
${\to x}_{c m} = \frac{m (a^i + b^j)}{3 m} = \frac{1}{3} (a^i + b^j)$

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Three particles, each of mass m, are placed at the corners of a right-angled triangle as shown in the figure. If OA = a and OB = b, the position vector of the centre of mass is

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a b c - (a + b + c) =

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Three particles, each of mass m are placed at the vertices of a right angled triangle as shown in figure. The position vector of the center of mass of the system is: (O is the origin and ^i,^j,^k are unit vectors)

The correct option is B 13(a^i+b^j)Position of centre of mass is given by, →xcm=∑mi→xi∑mi→xcm=m(a^i+b^j)3m=13(a^i+b^j)

Three particles, each of mass m are placed at the vertices of a right angled triangle as shown in figure. The position vector of the center of mass of the system is:
(O is the origin and $^i,^j,^k$ are unit vectors)

The correct option is B $\frac{1}{3} (a^i + b^j)$
Position of centre of mass is given by, ${\to x}_{c m} = \frac{\sum m_{i} {\to x}_{i}}{\sum m_{i}}$
${\to x}_{c m} = \frac{m (a^i + b^j)}{3 m} = \frac{1}{3} (a^i + b^j)$