For non-zero vectors bar{a}, bar{b}, bar{c},i | vedclass

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

Question

(D) Given $\bar{a} 	imes \bar{b} = \bar{c}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Substituting $\bar{c} = \bar{a} 	imes \bar{b}$ into the second eq

Vedclass · Accepted Answer

$|\bar{b}| = 1, |\bar{c}| = |\bar{a}|$

Vedclass · Answer

(D) Given $\bar{a} 	imes \bar{b} = \bar{c}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Substituting $\bar{c} = \bar{a} 	imes \bar{b}$ into the second equation: $\bar{b} 	imes (\bar{a} 	imes \bar{b}) = \bar{a}$.Using the vector triple product formula $\bar{b} 	imes (\bar{a} 	imes \bar{b}) = (\bar{b} \cdot \bar{b})\bar{a} - (\bar{b} \cdot \bar{a})\bar{b} = |\bar{b}|^2 \bar{a} - (\bar{a} \cdot \bar{b})\bar{b}$.Thus,$|\bar{b}|^2 \bar{a} - (\bar{a} \cdot \bar{b})\bar{b} = \bar{a}$.Since $\bar{b} 	imes \bar{c} = \bar{a}$,$\bar{b}$ is perpendicular to $\bar{a}$,so $\bar{a} \cdot \bar{b} = 0$.Substituting this,we get $(|\bar{b}|^2 - 1)\bar{a} = 0$. Since $\bar{a} 
eq 0$,$|\bar{b}|^2 = 1$,so $|\bar{b}| = 1$.From $\bar{a} 	imes \bar{b} = \bar{c}$,we have $|\bar{a} 	imes \bar{b}| = |\bar{c}|$.Since $\bar{a} \perp \bar{b}$,$|\bar{a} 	imes \bar{b}| = |\bar{a}||\bar{b}| \sin(90^\circ) = |\bar{a}|(1)(1) = |\bar{a}|$.Therefore,$|\bar{c}| = |\bar{a}|$. Thus,the correct option is $D$.

For non-zero vectors $\bar{a}, \bar{b}, \bar{c}$,if $\bar{a} \times \bar{b} = \bar{c}$ and $\bar{b} \times \bar{c} = \bar{a}$,then:

Similar Questions

If $A(3, 1, -1)$,$B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$,$C(2, 2, 1)$ and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are the vertices of a quadrilateral $ABCD$,then its area is

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$,then $x=$

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$,then ${(a+3b) \times (3a-b)}^2$ is equal to

For vectors $\bar{a}$ and $\bar{b}$,$|\bar{a}| = \frac{2}{3}$,$|\bar{b}| = 3$ and $|\bar{a} \times \bar{b}| = 1$,then the angle between $\bar{a}$ and $\bar{b}$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module