Let a,b and c be three non-zero vectors such that b and c are non-collinear.

Question

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c},\vec{b}+6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a}+\alpha \vec{b}+\beta \vec{c}=\overset{	o }{0}$, then $\alpha +\beta $ is equal to

Accepted Answer

$\vec a+5\vec b = \lambda\vec c$ \ldots  (1); $\vec b+6\vec c = \mu\vec a$ \ldots  (2).

From (2): $\vec a = \dfrac{\vec b+6\vec c}{\mu}$. Substituting into (1): $\dfrac{\vec b+6\vec c}{\mu}+5\vec b=\lambda\vec c$.

Comparing $\vec b$ coefficients: $\dfrac1\mu+5=0 \Rightarrow \mu=-\dfrac15$.

Comparing $\vec c$ coefficients: $\dfrac6\mu=\lambda \Rightarrow \lambda=-30$.

So $\vec a = \dfrac{\vec b+6\vec c}{-1/5} = -5\vec b-30\vec c$, i.e. $\vec a+5\vec b+30\vec c=\vec 0$.

Comparing with $\vec a+\alpha\vec b+\beta\vec c=\vec 0$: $\alpha=5,\ \beta=30 \Rightarrow \alpha+\beta=35$.

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