Question

Let $\vec a,\vec b$ and $\vec c$ be three non-zero vectors such that no two of them are collinear. If the vector $\vec a+\vec b$ is collinear with $\vec c$ and $\vec b+\vec c$ is collinear with $\vec a$, then \[ \vec a+\vec b+\vec c= \]

• A. $\vec a$
• B. $\vec b$
• C. $\vec c$
• D. $\vec 0$

Hint:

When sums of vectors are collinear with another vector, convert the condition into scalar-multiple equations.

Answer 1

The Correct Option is D

Step 1: Turn collinearity into equations.
$\vec a+\vec b$ collinear with $\vec c$ means $\vec a+\vec b=\lambda\vec c$ for some number $\lambda.$ Also $\vec b+\vec c$ collinear with $\vec a$ means $\vec b+\vec c=\mu\vec a.$
Step 2: Substitute one into the other.
From the first, $\vec a=\lambda\vec c-\vec b.$ Put this in the second: $\vec b+\vec c=\mu(\lambda\vec c-\vec b).$
Step 3: Group like vectors.
Rearranging: $\vec b+\mu\vec b+\vec c-\mu\lambda\vec c=0$, i.e. $(1+\mu)\vec b+(1-\mu\lambda)\vec c=\vec 0.$
Step 4: Use that no two are parallel.
Since $\vec b$ and $\vec c$ are not collinear, the only way their combination is zero is if both coefficients vanish: $1+\mu=0$ and $1-\mu\lambda=0.$
Step 5: Solve the coefficients.
So $\mu=-1$, then $1-(-1)\lambda=0$ gives $\lambda=-1.$ Hence $\vec a+\vec b=-\vec c.$
Step 6: Read off the answer.
$\vec a+\vec b=-\vec c$ rearranges to $\vec a+\vec b+\vec c=\vec0.$ \[ \boxed{\vec0} \]