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

The correct option is B λ = 94,μ = 2 Consider the given position vectors as O⃗A⃗, O⃗B⃗ , O⃗C⃗ respectivelyWe know that #160; A⃗B⃗=O⃗B⃗ O⃗ ...

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Standard XII

Physics

Vectors and Its Types

The position ...

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 = - \frac{9}{4}, μ = 2

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λ = \frac{9}{4}, μ = - 2

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none of these

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Solution

The correct option is B $λ = \frac{9}{4}, μ = - 2$
Consider the given position vectors as $\to O A, \to O B, \to O C$ respectively
We know that
$\to A B = \to O B - \to O A$

$\Rightarrow \to A B = (\to a - \to 2 b + λ \to c) - (2 \to a - \to b + 3 \to c)$
$\Rightarrow \to A B = - \to a - \to b + (λ - 3) \to c$
Similarly, we find $\to A C = \to O C - O A$ ; $\to B C = \to O C - \to O B$
$\Rightarrow \to A C = (μ - 2) \to a - 4 \to b - 3 \to c$
$\Rightarrow \to B C = (μ - 1) \to a + 4 \to b + 3 \to c$

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

On solving $(1)$ and $(2)$ we obtain $μ = - 2$ and $λ = \frac{9}{4}$

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Similar questions

Q. Assertion :Let

\to a, \to b, \to c

and

\to d

be the position vectors of four points

A, B, C

and

D

and

3 \to a - 2 \to b + 5 \to c - 6 \to d = \to 0

, then points

A, B, C

and

D

are coplanar. Reason: Three non-zero, linearly dependent coinitial vectors

(\to P Q, \to P R

and

\to P S)

are coplanar, then

\to P Q = λ \to P R + μ \to P S

, where

λ

and

μ

are scalars.

Q. Four points given by position vectors

2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c

are coplanar, where

\to a, \to b a n d \to c

are non-coplanar vectors.

Q. 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

Q. 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

Q. Assertion :Statement- 1: If the vectors

\to a

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\to c

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and

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The position vectors of three points are 2→a−→b+3→c,→a−2→b+λ→c and μ→a−5→b where →a,→b,→c are 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