CameraIcon

SearchIcon

MyQuestionIcon

1

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

Condition for Coplanarity of Four Points

Let a,b,c be ...

Question

Let $\vec{a}$ , $\vec{b}$ , $\vec{c}$ be three unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ . If $λ = \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ and $\vec{d} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ , then the ordered pair $(λ, \vec{d})$ is

A

$(\frac{3}{2}, 3 \vec{a} \times \vec{c})$

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

B

$(- \frac{3}{2}, 3 \vec{c} \times \vec{b})$

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

C

$(- \frac{3}{2}, 3 \vec{a} \times \vec{b})$

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

D

$(\frac{3}{2}, 3 \vec{b} \times \vec{c})$

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

Open in App

Solution

The correct option is C
$(- \frac{3}{2}, 3 \vec{a} \times \vec{b})$

Explanation for correct option:
Finding the ordered pair:
$a$ , $b$ , $c$ be three unit vectors. [Given]
$∴ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1$ .
Since, $\vec{a} + \vec{b} + \vec{c} = 0$ , we get,
$(\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = 0 \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{a} . \overset{⇀}{c} + \overset{⇀}{b} . \overset{⇀}{a} + \vec{b} . \vec{b} + \vec{b .} \overset{⇀}{c} + + \vec{c} . \vec{a} + \vec{c} . \vec{b} + \vec{c} . \vec{c} = 0 {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} . \vec{b} + \vec{b .} \vec{c} + \vec{c} . \vec{a}) = 0 1 + 1 + 1 + 2 (\vec{a} . \vec{b} + \vec{b .} \vec{c} + \vec{c} . \vec{a}) = 0 [∵ |a| = |b| = |c| = 1] 3 + 2 λ = 0 [∵ λ = \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}] λ = \frac{- 3}{2}$
Calculating the value of $\vec{d}$
$\vec{d} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{a} \times \vec{b} + \vec{b} \times (- \vec{a} - \vec{b}) + (- \vec{a} - \vec{b}) \times \vec{a} [∵ \vec{a} + \vec{b} + \vec{c} = 0] = \vec{a} \times \vec{b} + (\vec{b} \times - \vec{a}) - (\vec{b} \times \vec{b}) - (\vec{a} \times \vec{a}) + (- \vec{b} \times \vec{a}) = \vec{a} \times \vec{b} - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{a}) [∵ \vec{a} \times \vec{a} = 0] = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} [∵ \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})] = 3 (\vec{a} \times \vec{b})$
Therefore, the ordered pair $(λ, \vec{d})$ is $(- \frac{3}{2}, 3 (\vec{a} \times \vec{b})$ .
Hence, option (C) is the correct answer.

Suggest Corrections

thumbs-up

0

Similar questions

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

are non-zero vectors, then

[\to a \times \to b \to a \times \to c \to d]

can be simplified as:

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

are non-zero vectors satisfying

\to a = \to b + \to c

\to b \times \to d = \to 0

\to c \cdot \to d = 0

\frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}

is always equal to

\to a, \to b, \to c

be non-coplanar vectors and

\to d

be a non-zero vector which is perpendicular to

\to a + \to b + \to c

\to d = x \to b \times \to c + y \to c \times \to a + z \to a \times \to b

Q. For any three vectors

\to a, \to b

\to c

, prove that vectors

\to a - \to b, \to b - \to c

\to c - \to a

\to a

\to b

are two non zero vectors such that

\to a \times \to b = \to b \times \to a,

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Condition for Coplanarity of Four Points

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon