MyQuestionIcon

1

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

Collinear Vectors

Let a, b and ...

Question

Let $\to a, \to b$ and $\to c$ be three non -zero vectors such that they are mutually non collinear. If the vector $\to a + 2 \to b$ is collinear with $\to c$ and $\to b + 3 \to c$ is collinear with $\to a$ then $\to a + 2 \to b + 6 \to c$ equals

A

$λ \to a$ $(λ$ being some non -zero scalar)

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

B

$λ \to b$ $(λ$ being some non -zero scalar)

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

C

$λ \to c$ $(λ$ being some non -zero scalar)

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

D

$0$

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

Open in App

Solution

The correct option is D
$0$

This question is about the collinearity of vectors. As we learnt one very useful way of expressing that two vectors are collinear is by stating that one of those vectors can be equated to a scalar multiple of the second vector.
In the question we are given that $\to a + \to 2 b$ is collinear to $\to c$ . So expressing the first vector as a scalar multiple of the second we can write as,

$(\to a + 2 \to b) = t_{1} \to c . . . (1)$
Similarly for the second set of collinear vectors we can write,
and $\to b + 3 \to c = t_{2} \to a . . . (2)$
Note that $t_{1}$ and $t_{2}$ are the scalar used here. Now on solving the equations $(1)$ and $(2)$ ,
$(1) - 2 \times (2) \Rightarrow \to a (1 + 2 t_{2}) + \to c (- t_{1} - 6) = 0$
Look carefully, this equation is of the form,
$\to a \cdot (scalar 1) + \to c \cdot (scalar 2) = 0$ .
In addition to this we are given in the question that $\to a, \to b$ and $\to c$ be three non-zero vectors such that no two of these are collinear. Therefore it is safe to assume that scalars 1 and 2 are zero.
$\Rightarrow 1 + 2 t_{2} = 0 \Rightarrow t_{2} = - \frac{1}{2} & t_{1} = - 6$ .
Since $\to a$ and $\to c$ are non -collinear.
Putting the value of $t_{1}$ and $t_{2}$ in $(1)$ and $(2)$ , we get $\to a + 2 \to b + 6 \to c = \to 0$ .

Suggest Corrections

thumbs-up

0

Similar questions

a

b

c

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

a + 2 b

is collinear with

c

b + 3 c

is collinear with

a

a + 2 b + 6 c

a, b, c

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

a + 2 b

is collinear with

c

b + 3 c

is collinear with

a

λ

being some non-zero scalar), then

a + 2 b + 6 c

a, b

c

be three non-zero vectors, no two of which are collinear. If the vector

a + 2 b

is collinear with

c,

b + 3 c

is collinear with

a,

a + 2 b + 6 c

(λ

being some non-zero scalar) then

a + 2 b + 6 c =

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

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acceptor Circuit

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Collinear Vectors

Standard XIII Mathematics

Join BYJU'S Learning Program

CrossIcon