The vectors a = 2hat{i} + 3hat{j} + 6hat{k} a | vedclass

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(B) Given that vectors $a$ and $b$ are collinear,we can write $b = \lambda a$ for some scalar $\lambda$. Taking the magnitude on both sides,we have $|

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$\pm(6\hat{i} + 9\hat{j} + 18\hat{k})$

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(B) Given that vectors $a$ and $b$ are collinear,we can write $b = \lambda a$ for some scalar $\lambda$. Taking the magnitude on both sides,we have $|b| = |\lambda| |a|$. First,calculate the magnitude of vector $a$: $|a| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$. Given $|b| = 21$,substitute the values into the equation: $21 = |\lambda| 	imes 7$. $|\lambda| = \frac{21}{7} = 3$. Thus,$\lambda = \pm 3$. Substituting $\lambda$ back into the expression for $b$: $b = \pm 3(2\hat{i} + 3\hat{j} + 6\hat{k}) = \pm(6\hat{i} + 9\hat{j} + 18\hat{k})$.

The vectors $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $b$ are collinear and $|b| = 21$,then $b =$

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