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(A) Let the points be $A(\alpha, 10, 13)$,$B(6, 11, 11)$,and $C(\frac{9}{2}, \beta, -8)$.Since the points are collinear,the vectors $\vec{AB}$ and $\v

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) Let the points be $A(\alpha, 10, 13)$,$B(6, 11, 11)$,and $C(\frac{9}{2}, \beta, -8)$.Since the points are collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.$\vec{AB} = (6 - \alpha) \hat{i} + (11 - 10) \hat{j} + (11 - 13) \hat{k} = (6 - \alpha) \hat{i} + 1 \hat{j} - 2 \hat{k}$.$\vec{BC} = (\frac{9}{2} - 6) \hat{i} + (\beta - 11) \hat{j} + (-8 - 11) \hat{k} = -\frac{3}{2} \hat{i} + (\beta - 11) \hat{j} - 19 \hat{k}$.Since $\vec{AB} = k \vec{BC}$,we have:$\frac{6 - \alpha}{-3/2} = \frac{1}{\beta - 11} = \frac{-2}{-19} = \frac{2}{19}$.From $\frac{1}{\beta - 11} = \frac{2}{19}$,we get $2(\beta - 11) = 19 \implies 2\beta - 22 = 19 \implies 2\beta = 41$.From $\frac{6 - \alpha}{-3/2} = \frac{2}{19}$,we get $6 - \alpha = \frac{2}{19} 	imes (-\frac{3}{2}) = -\frac{3}{19} \implies \alpha = 6 + \frac{3}{19} = \frac{114 + 3}{19} = \frac{117}{19}$.Now,calculate $(19 \alpha - 6 \beta)^2 = (19 	imes \frac{117}{19} - 3 	imes 2\beta)^2 = (117 - 3 	imes 41)^2 = (117 - 123)^2 = (-6)^2 = 36$.

If the points with position vectors $\alpha \hat{i} + 10 \hat{j} + 13 \hat{k}$,$6 \hat{i} + 11 \hat{j} + 11 \hat{k}$,and $\frac{9}{2} \hat{i} + \beta \hat{j} - 8 \hat{k}$ are collinear,then $(19 \alpha - 6 \beta)^2$ is equal to $...........$.

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