If a - b - c are three non-zero vectors- no two of which are collinear and the vector a - b is collinear with c- and b - c is collinear with a- then a - b - c is equal to

If $\to a, \to b, \to c$ are three non-zero vectors, no two of which are collinear, $\to a + \to b$ is collinear with $\to c$ and $\to b + 3 \to c$ is collinear with $\to a$ , then $| \to a + 2 \to b + 6 \to c |$ will be equal to

Q. Let

a

b

and

c

be three non-zero vectors such that no two of these are collinear. if the vectors

a + 2 b

is collinear with

c

and

b + 3 c

is collinear with

a

.then

a + 2 b + 6 c

is equal to :

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If →a, →b, →c are three non-zero vectors, no two of which are collinear and the vector →a+→b is collinear with →c, and →b+→c is collinear with →a, then →a+→b+→c is equal to

The correct option is B →0Let a+b=xc and b+c=yaThen, a+b+c=(1+x)c=(1+y)aIf x≠−1, then c=1+y1+xa∴c and a are collinear which is contradiction∴x=−1,∴a+b+c=0

If $\to a, \to b, \to c$ are three non-zero vectors, no two of which are collinear and the vector $\to a + \to b$ is collinear with $\to c$ , and $\to b + \to c$ is collinear with $\to a$ , then $\to a + \to b + \to c$ is equal to

The correct option is B $\to 0$
Let $a + b = x c$ and $b + c = y a$
Then, $a + b + c = (1 + x) c = (1 + y) a$
If $x \neq - 1$ , then $c = \frac{1 + y}{1 + x} a$
$∴ c$ and $a$ are collinear which is contradiction
$∴ x = - 1, ∴ a + b + c = 0$