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Question

(A) Let the points be $A(\vec{a}+\vec{b})$, $B(\vec{b}+\vec{c})$, and $C(\vec{c}+\vec{a})$.The vectors lying on the plane are $\vec{AB} = B - A = (\ve

Vedclass · Accepted Answer

$\vec{r} \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = 2[\vec{a} \vec{b} \vec{c}]$

Vedclass · Answer

(A) Let the points be $A(\vec{a}+\vec{b})$, $B(\vec{b}+\vec{c})$, and $C(\vec{c}+\vec{a})$.The vectors lying on the plane are $\vec{AB} = B - A = (\vec{b}+\vec{c}) - (\vec{a}+\vec{b}) = \vec{c} - \vec{a}$ and $\vec{AC} = C - A = (\vec{c}+\vec{a}) - (\vec{a}+\vec{b}) = \vec{c} - \vec{b}$.The normal vector $\vec{n}$ to the plane is $\vec{AB} 	imes \vec{AC} = (\vec{c} - \vec{a}) 	imes (\vec{c} - \vec{b}) = \vec{c} 	imes \vec{c} - \vec{c} 	imes \vec{b} - \vec{a} 	imes \vec{c} + \vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a} + \vec{a} 	imes \vec{b}$.The equation of the plane is $\vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n}$.$\vec{A} \cdot \vec{n} = (\vec{a}+\vec{b}) \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = \vec{a} \cdot (\vec{b} 	imes \vec{c}) + \vec{b} \cdot (\vec{c} 	imes \vec{a}) = [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] = 2[\vec{a} \vec{b} \vec{c}]$.Thus, the equation is $\vec{r} \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = 2[\vec{a} \vec{b} \vec{c}]$.

The equation of the plane passing through the points having position vectors $\vec{a} + \vec{b}$, $\vec{b} + \vec{c}$ and $\vec{c} + \vec{a}$ is

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