જો bar{a}, bar{b}, bar{c} એ અસમતલીય સદિશો હોય | vedclass

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(C) આપેલ છે કે $\bar{p}=\frac{\bar{b} 	imes \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{q}=\frac{\bar{c} 	imes \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \b

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) આપેલ છે કે $\bar{p}=\frac{\bar{b} 	imes \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{q}=\frac{\bar{c} 	imes \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{r}=\frac{\bar{a} 	imes \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$.આપણે જાણીએ છીએ કે અદિશ ત્રિગુણિત ગુણાકાર $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$ થાય.હવે,દરેક પદની ગણતરી કરીએ:$\bar{a} \cdot \bar{p} = \bar{a} \cdot \left( \frac{\bar{b} 	imes \bar{c}}{[\bar{a} \bar{b} \bar{c}]} ight) = \frac{\bar{a} \cdot (\bar{b} 	imes \bar{c})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.તે જ રીતે,$\bar{b} \cdot \bar{q} = \bar{b} \cdot \left( \frac{\bar{c} 	imes \bar{a}}{[\bar{a} \bar{b} \bar{c}]} ight) = \frac{\bar{b} \cdot (\bar{c} 	imes \bar{a})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{b} \bar{c} \bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.અને $\bar{c} \cdot \bar{r} = \bar{c} \cdot \left( \frac{\bar{a} 	imes \bar{b}}{[\bar{a} \bar{b} \bar{c}]} ight) = \frac{\bar{c} \cdot (\bar{a} 	imes \bar{b})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{c} \bar{a} \bar{b}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.તેથી,$\bar{a} \cdot \bar{p} + \bar{b} \cdot \bar{q} + \bar{c} \cdot \bar{r} = 1 + 1 + 1 = 3$.

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