If bar{a}, bar{b}, bar{c} are three non-zero | vedclass

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(B) Given that $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,there exists a non-zero scalar $n$ such that $\bar{a}+2 \bar{b} = n \bar{c}$. (Equation

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$-6 \bar{c}$

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(B) Given that $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,there exists a non-zero scalar $n$ such that $\bar{a}+2 \bar{b} = n \bar{c}$. (Equation $1$)Similarly,since $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,there exists a non-zero scalar $m$ such that $\bar{b}+3 \bar{c} = m \bar{a}$. (Equation $2$)From Equation $2$,we have $\bar{b} = m \bar{a} - 3 \bar{c}$.Substitute this into Equation $1$: $\bar{a} + 2(m \bar{a} - 3 \bar{c}) = n \bar{c}$.This simplifies to: $\bar{a} + 2m \bar{a} - 6 \bar{c} = n \bar{c}$.$(1 + 2m) \bar{a} = (n + 6) \bar{c}$.Since $\bar{a}$ and $\bar{c}$ are non-zero and not collinear,the coefficients must be zero.Therefore,$1 + 2m = 0 \Rightarrow m = -\frac{1}{2}$ and $n + 6 = 0 \Rightarrow n = -6$.Substituting $n = -6$ into Equation $1$,we get $\bar{a} + 2 \bar{b} = -6 \bar{c}$.

If $\bar{a}, \bar{b}, \bar{c}$ are three non-zero vectors,no two of them are collinear,$\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,and $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,then $\bar{a}+2 \bar{b}$ is equal to:

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