If (vec{a} times vec{b}) times vec{c} = vec{a | vedclass

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

Question

(D) Using the vector triple product formula: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a

Vedclass · Accepted Answer

parallel

Vedclass · Answer

(D) Using the vector triple product formula: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.Similarly,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Equating the two expressions:$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Subtracting $(\vec{a} \cdot \vec{c})\vec{b}$ from both sides:$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$.This implies $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$.Since $\vec{a} \cdot \vec{b} 
eq 0$ and $\vec{b} \cdot \vec{c} 
eq 0$,we can write $\vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}} ight) \vec{c}$.This shows that $\vec{a}$ is a scalar multiple of $\vec{c}$,which means $\vec{a}$ and $\vec{c}$ are parallel.

If $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$ where $\vec{a}, \vec{b},$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{b} \cdot \vec{c} \neq 0$,then $\vec{a}$ and $\vec{c}$ are:

Similar Questions

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\bar{b} + \bar{c}}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is:

If $\vec{x} \cdot \vec{y} = 0$ then,$(\vec{y} \times \vec{x}) \times \vec{x} = $ . . . . . . . where,$|\vec{x}| = 1$.

Let $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+2 \hat{j}+3 \hat{k}$. Then the vector product $(\vec{a}+\vec{b}) \times((\vec{a} \times((\vec{a}-\vec{b}) \times \vec{b})) \times \vec{b})$ is equal to:

$\overrightarrow{a} \times [\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module