If a = alpha hat{i} + 3 hat{j} - 6 hat{k} and | vedclass

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

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$(-6, 2)$

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(A) Two vectors $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $b = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ 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 $a = \alpha \hat{i} + 3 \hat{j} - 6 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + \beta \hat{k}$.Comparing the components,we have:$\frac{\alpha}{2} = \frac{3}{-1} = \frac{-6}{\beta}$From $\frac{\alpha}{2} = -3$,we get $\alpha = -6$.From $\frac{3}{-1} = \frac{-6}{\beta}$,we get $-3\beta = -6$,which implies $\beta = 2$.Thus,the values are $\alpha = -6$ and $\beta = 2$.

If $a = \alpha \hat{i} + 3 \hat{j} - 6 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + \beta \hat{k}$,then the values of $\alpha, \beta$ so that $a$ and $b$ may be collinear are

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