Let a, b, c be three non-zero vectors which are pairwise non-collinear.

Question

Let $\vec a,\vec b,\vec c$ be three non-zero vectors which are pairwise non-collinear. If $\vec a+3\vec b$ is collinear with $\vec c$ and $\vec b+2\vec c$ is collinear with $\vec a$, then $\vec a+3\vec b+6\vec c$ is:

Accepted Answer

Let $\vec a+3\vec b=\lambda\vec c$ \ldots (i) and $\vec b+2\vec c=\mu\vec a$ \ldots (ii). From (ii), $\vec b=\mu\vec a-2\vec c$; substituting into (i):

$$\vec a+3\mu\vec a-6\vec c=\lambda\vec c \implies (1+3\mu)\vec a=(\lambda+6)\vec c$$

Since $\vec a,\vec c$ are non-collinear, $1+3\mu=0$ and $\lambda+6=0$, so $\lambda=-6$. From (i): $\vec a+3\vec b=-6\vec c \implies \vec a+3\vec b+6\vec c=\vec 0$

Solutions · 1