If the vectors a, b and c form the sides be the sides AB, CA, and AB respectively of a triangle ABC, then

\to a . \to b + \to b . \to c + \to c . \to a = 0

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

\to a \times \to b = \to b \times \to c = \to c \times \to a

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

\to a . \to b = \to b . \to c = \to c . \to a

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

\to a \times \to b + \to b \times \to c + \to c \times \to a = 0

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

Open in App

Solution

The correct option is B $\to a \times \to b = \to b \times \to c = \to c \times \to a$
Consider the problem
For a triangle
$\to a + \to b + \to c = 0$ ----- (i)
Take cross product of $\to b$ on both sides

$\to a \times \to a + \to a \times \to b + \to a \times \to c = 0$

$\to b \times \to a + \to c \times \to a = 0$

$\to c \times \to a = \to a \times \to b$ ---- (ii)

Similarly take cross product of $\to b$ both side in equation (i)

$\begin{matrix} \to a \times \to b + \to b \times \to b + \to c \times \to b = 0 \to a \times \to b = \to b \times \to c - - - - (i i i) \end{matrix}$
from (ii) and (iii)

$\to a \times \to b = \to b \times \to c = \to c \times \to a$

Hence option $B$ is correct answer.

Suggest Corrections

Similar questions

[\to a \to b \to c] = 2

\frac{\to a . (\to b \times \to c)}{(\to c \times \to a) . \to b} + \frac{\to b . (\to c \times \to a)}{(\to a \times \to b) . \to c} + \frac{\to c . (\to a \times \to b)}{(\to b \times \to c) . \to a} =

[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 1

\frac{\to a . \to b \times \to c}{\to c \times \to a . \to b} + \frac{\to b . \to c \times \to a}{\to a \times \to b . \to c} + \frac{\to c . \to a \times \to b}{\to b \times \to c . \to a}

\to a, \to b, \to c

\to a + \to b + \to c = 0

\to a . \to b + \to b . \to c + \to c . \to a

\to a, \to b, \to c

\to a + \to b + \to c = 0

μ = \to a . \to b + \to b . \to c + \to c . \to a

| \to a | = 1; ∣ ∣ \to b ∣ ∣ = 4; | \to c | = 2

| \to a |, ∣ ∣ \to b ∣ ∣

| \to c |

\to a + \to b + \to c = 0

λ = \to a . \to b + \to b . \to c + \to c . \to a

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 4 a n d | \to c | = 2

Join BYJU'S Learning Program