If vectors overrightarrow{a}_{1} = x hat{i} - | vedclass

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(D) Since $\overrightarrow{a}_{1}$ and $\overrightarrow{a}_{2}$ are collinear,their components must be proportional: $\frac{x}{1} = \frac{-1}{y} = \fr

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$\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$

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(D) Since $\overrightarrow{a}_{1}$ and $\overrightarrow{a}_{2}$ are collinear,their components must be proportional: $\frac{x}{1} = \frac{-1}{y} = \frac{1}{z} = k$ (where $k$ is a constant). From this,we get $x = k$,$y = -\frac{1}{k}$,and $z = \frac{1}{k}$. Substituting these into the vector $x \hat{i} + y \hat{j} + z \hat{k}$,we get $k \hat{i} - \frac{1}{k} \hat{j} + \frac{1}{k} \hat{k}$. For the vectors to be collinear,the ratio must hold. If we consider the simplest case where $k=1$,then $x=1, y=-1, z=1$. The vector becomes $\hat{i} - \hat{j} + \hat{k}$. The magnitude of this vector is $\sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$. The unit vector is $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.

If vectors $\overrightarrow{a}_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{a}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear,then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is ...... .

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