If a is collinear with b = 3 hat{i} + 6 hat{j | vedclass

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(C) Given that $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$.Therefore,$a = \lambda b$ for some scalar $\lambda$.We are given the dot

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$3$

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(C) Given that $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$.Therefore,$a = \lambda b$ for some scalar $\lambda$.We are given the dot product $a \cdot b = 27$.Substituting $a = \lambda b$ into the dot product equation:$(\lambda b) \cdot b = 27$$\lambda (b \cdot b) = 27$$\lambda |b|^2 = 27$First,calculate $|b|^2$:$|b|^2 = (3)^2 + (6)^2 + (6)^2 = 9 + 36 + 36 = 81$.Now,substitute this back into the equation:$\lambda (81) = 27$$\lambda = \frac{27}{81} = \frac{1}{3}$.Since $a = \lambda b$,we have $|a| = |\lambda b| = |\lambda| |b|$.Calculate $|b| = \sqrt{81} = 9$.Thus,$|a| = |\frac{1}{3}| 	imes 9 = 3$.

If $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$ and $a \cdot b = 27$,then $|a| =$

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