Question

If vector $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c = 0$, then vector $\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a$ is equal to
$A)6(\overrightarrow b \times \overrightarrow c)$
$B)3(\overrightarrow b \times \overrightarrow c)$
$C)2(\overrightarrow b \times \overrightarrow c)$
$D)0$

Answer 1

First, we need to know about the concept of vectors.
The length of the straight line denotes the magnitude of the vector and the arrowhead gives its direction and thus the vector has both magnitude and direction.
But in the scalar, there is the only magnitude, which is the difference.
Vector can be represented as $\overrightarrow a$with the direction symbol
Formula used:
$\overrightarrow m \times \overrightarrow n = - (\overrightarrow n \times \overrightarrow m )$ (changing the order will change the sign)
$\overrightarrow m \times \overrightarrow m = 0$ (cross product of the same vector will be zero)

Complete step by step answer:
Since from the given that vector we have $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c = 0$ and we need to find the value of the vector $\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a$.
Now let us simplify the required value, $\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a \Rightarrow \overrightarrow b \times \overrightarrow c + \overrightarrow a \times \overrightarrow b + \overrightarrow c \times \overrightarrow a$
Since we know that $\overrightarrow m \times \overrightarrow n = - (\overrightarrow n \times \overrightarrow m )$and apply this in the last vector, thus we get $\overrightarrow b \times \overrightarrow c + \overrightarrow a \times \overrightarrow b + \overrightarrow c \times \overrightarrow a = \overrightarrow b \times \overrightarrow c + \overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c$ as $- \overrightarrow a \times \overrightarrow c = \overrightarrow c \times \overrightarrow a$
Taking out the common values we have, $\overrightarrow b \times \overrightarrow c + \overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c \Rightarrow \overrightarrow b \times \overrightarrow c + \overrightarrow a \times (\overrightarrow b - \overrightarrow c )$
Thus, after simplifying we have the vector as $\overrightarrow b \times \overrightarrow c + \overrightarrow a \times (\overrightarrow b - \overrightarrow c )$
Now from the given that we have $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c = 0$
Let us convert the given value as $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c = 0 \Rightarrow \overrightarrow a = - (2\overrightarrow b + 3\overrightarrow c )$ then substitute this value in the above-simplified value, that is $\overrightarrow b \times \overrightarrow c + \overrightarrow a \times (\overrightarrow b - \overrightarrow c )$then we get $\overrightarrow b \times \overrightarrow c + \overrightarrow a \times (\overrightarrow b - \overrightarrow c ) \Rightarrow \overrightarrow b \times \overrightarrow c - [(2\overrightarrow b + 3\overrightarrow c ) \times (\overrightarrow b - \overrightarrow c )]$
Now by the multiplication operation, we have $\overrightarrow b \times \overrightarrow c - [(2\overrightarrow b + 3\overrightarrow c ) \times (\overrightarrow b - \overrightarrow c )] \Rightarrow \overrightarrow b \times \overrightarrow c - [2\overrightarrow b \times \overrightarrow b - 2\overrightarrow b \times \overrightarrow c + 3\overrightarrow c \times \overrightarrow b - 3\overrightarrow c \times \overrightarrow c ]$
Applying the two conditions $\overrightarrow m \times \overrightarrow n = - (\overrightarrow n \times \overrightarrow m )$ (changing the order will change the sign) and $\overrightarrow m \times \overrightarrow m = 0$ (cross product of the same vector will be zero).
Thus, we have $\overrightarrow b \times \overrightarrow c - [2\overrightarrow b \times \overrightarrow b - 2\overrightarrow b \times \overrightarrow c + 3\overrightarrow c \times \overrightarrow b - 3\overrightarrow c \times \overrightarrow c ] \Rightarrow \overrightarrow b \times \overrightarrow c - [0 - 2\overrightarrow b \times \overrightarrow c + 3\overrightarrow c \times \overrightarrow b - 0]$
Changing the order of the variable $- \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow b$then we get $\overrightarrow b \times \overrightarrow c - [ - 2\overrightarrow b \times \overrightarrow c + 3\overrightarrow c \times \overrightarrow b ] \Rightarrow \overrightarrow b \times \overrightarrow c - [ - 2\overrightarrow b \times \overrightarrow c - 3\overrightarrow b \times \overrightarrow c ]$
Further solving the values, we get $\overrightarrow b \times \overrightarrow c - [ - 5\overrightarrow b \times \overrightarrow c ] \Rightarrow \overrightarrow b \times \overrightarrow c + 5\overrightarrow b \times \overrightarrow c \Rightarrow 6(\overrightarrow b \times \overrightarrow c )$

So, the correct answer is “Option A”.

Note: We just solved the given problem using the properties of the vector with the cross product.
The properties of the cross-product vector are not the same as the dot product vector, so these properties used will not be repeated if the problem is in the dot product.
The cross product represents as $\times$