If the vectors 2 hat{i}-3 hat{j}+6 hat{k} and | vedclass

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(C) Let $\vec{a} = 2 \hat{i}-3 \hat{j}+6 \hat{k}$.Since $\vec{b}$ is collinear to $\vec{a}$,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $

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$4 \hat{i}-6 \hat{j}+12 \hat{k}$

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(C) Let $\vec{a} = 2 \hat{i}-3 \hat{j}+6 \hat{k}$.Since $\vec{b}$ is collinear to $\vec{a}$,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda$.The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}$.Given $|\vec{b}| = 14$,the vector $\vec{b}$ can be expressed as $\vec{b} = \pm |\vec{b}| \hat{a}$.Taking the positive direction,$\vec{b} = 14 \left( \frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7} \right) = 2(2 \hat{i}-3 \hat{j}+6 \hat{k}) = 4 \hat{i}-6 \hat{j}+12 \hat{k}$.

If the vectors $2 \hat{i}-3 \hat{j}+6 \hat{k}$ and $\vec{b}$ are collinear and $|\vec{b}|=14$,then $\vec{b}$ has the value

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