Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.
(N/A) Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{b} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k}$.
We observe that $\vec{b} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k}$.
Taking $-2$ as a common factor,we get $\vec{b} = -2(2 \hat{i} - 3 \hat{j} + 4 \hat{k})$.
This can be written as $\vec{b} = -2 \vec{a}$.
Since $\vec{b} = \lambda \vec{a}$,where $\lambda = -2$,the two vectors are scalar multiples of each other.
Therefore,the given vectors are collinear.
We observe that $\vec{b} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k}$.
Taking $-2$ as a common factor,we get $\vec{b} = -2(2 \hat{i} - 3 \hat{j} + 4 \hat{k})$.
This can be written as $\vec{b} = -2 \vec{a}$.
Since $\vec{b} = \lambda \vec{a}$,where $\lambda = -2$,the two vectors are scalar multiples of each other.
Therefore,the given vectors are collinear.
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