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Let $\to a, \to b$ and $\to c$ be three non-zero vectors such that no two of these are collinear. If the vector $\to a + 2 \to b$ is collinear with $\to c$ and $\to b + 3 \to c$ is collinear with $\to a$ ( $λ$ being some non-zero scalar) then $\to a + 2 \to b + 6 \to c$ equals

A

λ ¯ ¯ ¯ a

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B

λ ¯ ¯ b

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C

λ ¯ ¯ c

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D

¯ ¯ ¯ 0

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Solution

The correct option is D $¯ ¯ ¯ 0$
$\to a + 2 \to b$ is collinear with $\to c$ .

So we can write:
$\to a + 2 \to b = t_{1} \to c$ ....... $(i)$
Similarly, $\to b + 3 \to c$ is collinear with $\to a$ .

So we can write $\to b + 3 \to c = t_{2} \to a$ ......... $(i i)$
To eliminate $\to b$ we will perform $(i) - 2 (i i)$ to get

$\to a - 6 \to c = t_{1} \to c - 2 t_{2} \to a$

$(1 + 2 t_{2}) \to a - (6 + t_{1}) \to c = 0$
Since $\to a$ and $\to c$ are non-collinear, both $1 + 2 t_{2}$ and $6 + t_{1}$ must be zero to satisfy above equation.
$∴ t_{2} = - \frac{1}{2}$ and $t_{1} = - 6$
Substituting the values of $t_{2}$ and $t_{1}$ in (i) we get
$\to a + 2 \to b + 6 \to c = \to 0$

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