The position vectors of three points are2a-b-3c- a-2b-c and -a-5b where a-b-care non-coplanar vectors- The points are collinear when

Consider the given position vectors as OA, OB, OC We know that, AB=OB OA AB=(a 2b+λ c) (2a b+3c) ⇒AB= a b+(λ 3)c Similarly, nbsp;AC=OC OA and BC=OC OB ⇒AC=(μ ...

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Question

The position vectors of three points are $2 \to a - \to b + 3 \to c,$ $\to a - 2 \to b + λ \to c$ and $μ \to a - 5 \to b$ where $\to a, \to b, \to c$ are non-coplanar vectors. The points are collinear when

λ = 2, μ = - \frac{9}{4}

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λ = - 2, μ = \frac{9}{4}

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λ = \frac{9}{4}, μ = - 2

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λ = \frac{9}{4}, μ = 2

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Solution

The correct option is C $λ = \frac{9}{4}, μ = - 2$
Consider the given position vectors are $- - \to O A, - - \to O B, - - \to O C$
We know that, $- - \to A B = - - \to O B - - - \to O A$
$- - \to A B = (\to a - \to 2 b + λ \to c) - (- \to 2 a - \to b + \to 3 c)$
$- - \to A B = - \to a - \to b + (λ - 3) \to c$
Similarly, we find
$- - \to A C = - - \to O C - - - \to O A$
$- - \to B C = - - \to O C - - - \to O B$
$- - \to A C = (μ - 2) \to a - \to 4 b - \to 3 c$
$- - \to B C = (μ - 1) \to a - \to 3 b - λ \to c$

We know that the required condition for vectors to be collinear is $- - \to A B | | - - \to A C$
Considering $- - \to A B$ and $- - \to A C$ we can write,
$\frac{- 1}{μ - 2} = \frac{- 1}{- 4}$
$\Rightarrow μ - 2 = - 4$
$\Rightarrow μ = - 2$
and $\frac{λ - 3}{- 3} = \frac{- 1}{- 4}$
$\Rightarrow λ - 3 = \frac{- 3}{4}$
$\Rightarrow λ = \frac{9}{4}$

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The position vectors of three points are2→a−→b+3→c, →a−2→b+λ→c and μ→a−5→b where →a,→b,→care non-coplanar vectors. The points are collinear when

The position vectors of three points are $2 \to a - \to b + 3 \to c,$ $\to a - 2 \to b + λ \to c$ and $μ \to a - 5 \to b$ where $\to a, \to b, \to c$ are non-coplanar vectors. The points are collinear when