# Two particles of mass 1 kg and 3 kg have position vectors 2 i + 3 j + 4 k and -2 i + 3 j - 4 k respectively. The centre of mass has a position vector (A) i + 3 j - 2 k B - i - 3 j - 2 k (c) - i + 3 j + 2 k D - i + 3 j - 2 k

## Question

Two particles of mass 1 kg and 3 kg have position vectors $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $-2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ respectively. The centre of mass has a position vector (A) $\hat{i} + 3 \hat{j} - 2 \hat{k}$ B $- \hat{i} - 3 \hat{j} - 2 \hat{k}$ (c) $- \hat{i} + 3 \hat{j} + 2 \hat{k}$ D $- \hat{i} + 3 \hat{j} - 2 \hat{k}$

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## Answer

**:**

To find the position vector of the center of mass of the two particles, we use the formula:

$$ \vec{r}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} $$

Given:

- Mass of the first particle, $m_1 = 1 \text{ kg}$
- Position vector of the first particle, $\vec{r}_1 = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
- Mass of the second particle, $m_2 = 3 \text{ kg}$
- Position vector of the second particle, $\vec{r}_2 = -2 \hat{i} + 3 \hat{j} - 4 \hat{k}$

Substitute these values into the formula:

[ \vec{r}_{\text{cm}} = \frac{1 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + 3 (-2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{1 + 3} ]

Expanding the terms in the numerator:

[ \vec{r}_{\text{cm}} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + 3(-2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{4} ]

Simplify the expression inside the numerator:

[ \vec{r}_{\text{cm}} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (-6 \hat{i} + 9 \hat{j} - 12 \hat{k})}{4} ]

Combine the like terms:

[ \vec{r}_{\text{cm}} = \frac{(2 \hat{i} - 6 \hat{i}) + (3 \hat{j} + 9 \hat{j}) + (4 \hat{k} - 12 \hat{k})}{4} ] [ \vec{r}_{\text{cm}} = \frac{(-4 \hat{i} + 12 \hat{j} - 8 \hat{k})}{4} ]

Now, divide each component by 4:

[ \vec{r}_{\text{cm}} = -\hat{i} + 3 \hat{j} - 2 \hat{k} ]

Thus, the position vector of the center of mass is:

**$\boxed{-\hat{i} + 3 \hat{j} - 2 \hat{k}}$**

Therefore, the correct option is **D**.

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