Step 1: {Formula for center of mass}
\[ \vec{r}_{{com}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]
Step 2: {Substituting values}
\[ \vec{r}_{{com}} = \frac{(1)(\hat{i} + 2\hat{j} + \hat{k}) + (3)(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} \] \[ = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4} \] \[ = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} \] \[ = -2\hat{i} - \hat{j} + \hat{k} \]
Step 3: {Find the magnitude}
\[ |\vec{r}_{{com}}| = \sqrt{(-2)^2 + (-1)^2 + (1)^2} \] \[ = \sqrt{4 + 1 + 1} = \sqrt{6} \] The only vector with the same magnitude is \( \hat{i} - 2\hat{j} + \hat{k} \).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
Arrange the following in the ascending order of wavelength (\( \lambda \)):
(A) Microwaves (\( \lambda_1 \))
(B) Ultraviolet rays (\( \lambda_2 \))
(C) Infrared rays (\( \lambda_3 \))
(D) X-rays (\( \lambda_4 \))
Choose the most appropriate answer from the options given below: