Let vec{a}, vec{b}, vec{c} be three vectors i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} = \vec{v} 
eq 0$.From $\vec{a} 	imes \vec{b} = \vec{b} 	imes \v

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} = \vec{v} 
eq 0$.From $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c}$,we have $(\vec{a} - \vec{c}) 	imes \vec{b} = 0$,which implies $\vec{a} - \vec{c} = k_1 \vec{b}$ for some scalar $k_1$.Similarly,from $\vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$,we have $(\vec{b} - \vec{a}) 	imes \vec{c} = 0$,which implies $\vec{b} - \vec{a} = k_2 \vec{c}$ for some scalar $k_2$.Adding these equations: $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ is not necessarily true,but rather $\vec{a} + \vec{b} + \vec{c}$ must be a vector such that the cross products are equal.Actually,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = \vec{k}$ where $\vec{k}$ is a constant vector.However,the condition $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$ implies $\vec{a} + \vec{b} + \vec{c} = 0$ is not required; instead,it implies that $\vec{a}, \vec{b}, \vec{c}$ are coplanar or satisfy specific relations.Given the symmetry,the only solution for the centroid $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$ under the constraint $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} 
eq 0$ is that there is only $1$ possible position for the centroid,which is the origin $\vec{0}$.

Similar Questions

Let $a, b, c$ be three non-coplanar vectors such that $r_1 = a - b + c$,$r_2 = b + c - a$,$r_3 = c + a + b$,and $r = 2a - 3b + 4c$. If $r = \lambda_1 r_1 + \lambda_2 r_2 + \lambda_3 r_3$,then:

If vectors $\vec{a}$ and $\vec{b}$ represent the sides $\vec{AB}$ and $\vec{BC}$ of a regular hexagon $ABCDEF$,find the vector represented by $\vec{FA}$.

The vector $\frac{1}{3}(2i - 2j + k)$ is

If $P$ is the point of intersection of the diagonals of a parallelogram $ABCD$ and $O$ is any point,then $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = .......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module