If hat{i}+2 hat{j}+3 hat{k} and 3 hat{i}+2 ha | vedclass

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(A) Let the sides of the parallelogram be represented by vectors $\vec{a} = \hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$.T

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$\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

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(A) Let the sides of the parallelogram be represented by vectors $\vec{a} = \hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$.The diagonals of the parallelogram are given by $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$.Calculating $\vec{d_1} = (\hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k}) = 4\hat{i} + 4\hat{j} + 4\hat{k} = 4(\hat{i} + \hat{j} + \hat{k})$.The unit vector parallel to $\vec{d_1}$ is $\frac{\vec{d_1}}{|\vec{d_1}|} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{4^2 + 4^2 + 4^2}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{16+16+16}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{48}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{4\sqrt{3}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.Calculating $\vec{d_2} = (\hat{i}+2 \hat{j}+3 \hat{k}) - (3 \hat{i}+2 \hat{j}+\hat{k}) = -2\hat{i} + 0\hat{j} + 2\hat{k} = -2\hat{i} + 2\hat{k}$.The unit vector parallel to $\vec{d_2}$ is $\frac{-2\hat{i} + 2\hat{k}}{\sqrt{(-2)^2 + 2^2}} = \frac{-2\hat{i} + 2\hat{k}}{\sqrt{8}} = \frac{-2\hat{i} + 2\hat{k}}{2\sqrt{2}} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.Comparing with the given options, the correct unit vector is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

If $\hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}+2 \hat{j}+\hat{k}$ are sides of a parallelogram, then a unit vector parallel to one of the diagonals of the parallelogram is

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