If the vectors x hat{i}-3 hat{j}+7 hat{k} and | vedclass

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(B) Two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ are collinear if their com

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$\frac{-9}{7}$

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(B) Two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ are collinear if their components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = k$. Given vectors are $x \hat{i}-3 \hat{j}+7 \hat{k}$ and $\hat{i}+y \hat{j}-z \hat{k}$. Therefore,$\frac{x}{1} = \frac{-3}{y} = \frac{7}{-z} = k$. From this,we get $x = k$,$y = -\frac{3}{k}$,and $z = -\frac{7}{k}$. Now,substitute these values into the expression $\frac{x y^2}{z}$: $\frac{x y^2}{z} = \frac{k \cdot (-\frac{3}{k})^2}{-\frac{7}{k}} = \frac{k \cdot \frac{9}{k^2}}{-\frac{7}{k}} = \frac{\frac{9}{k}}{-\frac{7}{k}} = -\frac{9}{7}$.

If the vectors $x \hat{i}-3 \hat{j}+7 \hat{k}$ and $\hat{i}+y \hat{j}-z \hat{k}$ are collinear,then the value of $\frac{x y^2}{z}$ is equal to:

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