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

The correct option is C λ =94, μ = 2 When points x, y, z are collinear, we have #160; α x + β y = (α + β)z Similarly, x(2a⃗ #160;b⃗ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Fundamental Theorem in 2D

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, then the points are collinear when

λ = - 2, μ = \frac{9}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

λ = - \frac{9}{4}, μ = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

λ = \frac{9}{4}, μ = - 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $λ = \frac{9}{4}, μ = - 2$
When points $x, y, z$ are collinear, we have $α x + β y = (α + β) z$
Similarly, $x (2 \to a - \to b + 3 \to c) + y (\to a - 2 \to b + λ \to c) = (x + y) (μ \to a - 5 \to b)$
$\Rightarrow$ comparing the coefficients of $\to a \to 2 x + y = x μ + y μ$
$\to b \to - x - 2 y = - 5 x - 5 y$
$\to c \to 3 x + λ y = 0$
$\Rightarrow 4 x = - 3 y$ and so $λ = \frac{9}{4}$
Also, $μ = - 2$

Suggest Corrections

Similar questions

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. 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. If

\to a, \to b

and

\to c

are unit vectors satisfying

| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 9

, then

| 2 \to a + 5 \to b + 5 \to c |

Q. If

\to a, \to b, \to c

are non-coplanar vectors, then show that the four points

2 \to a + \to b, \to a + 2 \to b + \to c, 4 \to a - 2 \to b - \to c

and

3 \to a + 4 \to b - 5 \to c

are coplanar.

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

Join BYJU'S Learning Program

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, then 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, then the points are collinear when