જો alpha hat{i} + 10 hat{j} + 13 hat{k},6 hat | vedclass

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Question

(A) ધારો કે બિંદુઓ $A(\alpha, 10, 13)$,$B(6, 11, 11)$,અને $C(\frac{9}{2}, \beta, -8)$ છે.બિંદુઓ સમરેખ હોવાથી,સદિશો $\vec{AB}$ અને $\vec{BC}$ સમાંતર હો

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) ધારો કે બિંદુઓ $A(\alpha, 10, 13)$,$B(6, 11, 11)$,અને $C(\frac{9}{2}, \beta, -8)$ છે.બિંદુઓ સમરેખ હોવાથી,સદિશો $\vec{AB}$ અને $\vec{BC}$ સમાંતર હોવા જોઈએ.$\vec{AB} = (6 - \alpha) \hat{i} + (11 - 10) \hat{j} + (11 - 13) \hat{k} = (6 - \alpha) \hat{i} + 1 \hat{j} - 2 \hat{k}$.$\vec{BC} = (\frac{9}{2} - 6) \hat{i} + (\beta - 11) \hat{j} + (-8 - 11) \hat{k} = -\frac{3}{2} \hat{i} + (\beta - 11) \hat{j} - 19 \hat{k}$.$\vec{AB} = k \vec{BC}$ હોવાથી:$\frac{6 - \alpha}{-3/2} = \frac{1}{\beta - 11} = \frac{-2}{-19} = \frac{2}{19}$.$\frac{1}{\beta - 11} = \frac{2}{19}$ પરથી,$2(\beta - 11) = 19 \implies 2\beta - 22 = 19 \implies 2\beta = 41$.$\frac{6 - \alpha}{-3/2} = \frac{2}{19}$ પરથી,$6 - \alpha = \frac{2}{19} 	imes (-\frac{3}{2}) = -\frac{3}{19} \implies \alpha = 6 + \frac{3}{19} = \frac{117}{19}$.હવે,$(19 \alpha - 6 \beta)^2 = (19 	imes \frac{117}{19} - 3 	imes 2\beta)^2 = (117 - 3 	imes 41)^2 = (117 - 123)^2 = (-6)^2 = 36$.

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