Statement (A) : If vec{a} is perpendicular to | vedclass

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

Vedclass · Accepted Answer

$(A)$ is true but $(R)$ is false.

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(C) Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.Since $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$.Substituting these values: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (0)\vec{b} - (0)\vec{c} = 0$.Thus,Statement $(A)$ is true.Regarding Reason $(R)$,if $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} 	imes \vec{c} = |\vec{b}||\vec{c}| \sin(90^\circ) \hat{n} = |\vec{b}||\vec{c}| \hat{n} 
eq 0$ (unless one of them is a zero vector).Therefore,Reason $(R)$ is false.

Statement $(A)$ : If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,then $\vec{a} \times (\vec{b} \times \vec{c}) = 0$.
Reason $(R)$ : If $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} \times \vec{c} = 0$.

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Statement $(A)$ : If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,then $\vec{a} \times (\vec{b} \times \vec{c}) = 0$.Reason $(R)$ : If $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} \times \vec{c} = 0$.