MyQuestionIcon

1

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

Test for Collinearity of Vectors

Let a and b b...

Question

Let $\to a$ and $\to b$ be two non-zero and non-collinear vectors. Then which of the following is/are always CORRECT?

A

\to a \times \to b = [\to a \to b^i]^i + [\to a \to b^j]^j + [\to a \to b^k]^k

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

B

\to a \cdot \to b = (\to a \cdot^i) (\to b \cdot^i) + (\to a \cdot^j) (\to b \cdot^j) + (\to a \cdot^k) (\to b \cdot^k)

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

C

If

\to u =^a - (^a \cdot^b)^b

and

\to v =^a \times^b,

then

∣ ∣ \to u ∣ ∣ = ∣ ∣ \to v ∣ ∣

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

D

If

\to c = \to a \times (\to a \times \to b)

and

\to d = \to b \times (\to a \times \to b),

then

\to c + \to d = \to 0

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

Open in App

Solution

The correct option is C If $\to u =^a - (^a \cdot^b)^b$ and $\to v =^a \times^b,$ then $∣ ∣ \to u ∣ ∣ = ∣ ∣ \to v ∣ ∣$
For any vector $\to r,$
$\to r = (\to r \cdot^i)^i + (\to r \cdot^j)^j + (\to r \cdot^k)^k$
Put $\to r = (\to a \times \to b)$
Hence, $\to a \times \to b = [\to a \to b^i]^i + [\to a \to b^j]^j + [\to a \to b^k]^k$

Let $\to a = (\to a \cdot^i)^i + (\to a \cdot^j)^j + (\to a \cdot^k)^k$
$\to b = (\to b \cdot^i)^i + (\to b \cdot^j)^j + (\to b \cdot^k)^k$
From above equations, we get
$\to a \cdot \to b = (\to a \cdot^i) (\to b \cdot^i) + (\to a \cdot^j) (\to b \cdot^j) + (\to a \cdot^k) (\to b \cdot^k)$

$\to u =^a - (^a \cdot^b)^b {∣ ∣ \to u ∣ ∣}^{2} = {∣ ∣^a - (^a \cdot^b)^b ∣ ∣}^{2} = {|^a |}^{2} + {∣ ∣ (^a \cdot^b)^b ∣ ∣}^{2} - 2 {(^a \cdot^b)}^{2} = 1 + 1 \cdot 1 \cdot {cos}^{2} θ - 2 {cos}^{2} θ = {sin}^{2} θ ∴ | \to u | = sin θ$
$∣ ∣ \to v ∣ ∣ = ∣ ∣^a \times^b ∣ ∣ = |^a | ∣ ∣^b ∣ ∣ sin θ = sin θ = ∣ ∣ \to u ∣ ∣$

$\to c + \to d = (\to a + \to b) \times (\to a \times \to b)$
$= ((\to a + \to b) \cdot \to b) \to a - ((\to a + \to b) \cdot \to a) \to b = (\to a \cdot \to b + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}) \to a - ({∣ ∣ \to a ∣ ∣}^{2} + \to b \cdot \to a) \to b \neq \to 0$
[ As if $\to c + \to d = \to 0,$ then $\to a$ and $\to b$ will become collinear. ]

Suggest Corrections

thumbs-up

3

Similar questions

\to a, \to b

\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

\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

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 =

\to a, \to b, \to c

be three non zero vectors which are pairwise non-collinear. If

\to a + 3 \to b

is collinear and

\to b + 2 \to c

is collinear with

\to a

\to a + 3 \to b + 6 \to 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. lf the vector

¯ ¯ ¯ a + 2 ¯ ¯ b

is collinear with

¯ ¯ c

¯ ¯ b + 3 ¯ ¯ c

is collinear with

¯ ¯ ¯ a

¯ ¯ ¯ a + ¯ ¯ b + 3 ¯ ¯ c =

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Test for Collinearity of 3 Points or 2 Vectors

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Test for Collinearity of Vectors

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon