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(B) 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

$36$

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(B) 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 proportional.$\vec{AB} = (6-\alpha)\hat{i} + (11-10)\hat{j} + (11-13)\hat{k} = (6-\alpha)\hat{i} + 1\hat{j} - 2\hat{k}$.$\vec{BC} = (\frac{9}{2}-6)\hat{i} + (\beta-11)\hat{j} + (-8-11)\hat{k} = -\frac{3}{2}\hat{i} + (\beta-11)\hat{j} - 19\hat{k}$.For collinearity,$\frac{6-\alpha}{-3/2} = \frac{1}{\beta-11} = \frac{-2}{-19}$.From $\frac{1}{\beta-11} = \frac{2}{19}$,we get $2(\beta-11) = 19 \Rightarrow 2\beta - 22 = 19 \Rightarrow 2\beta = 41 \Rightarrow 6\beta = 123$.From $\frac{6-\alpha}{-3/2} = \frac{2}{19}$,we get $19(6-\alpha) = -3 \Rightarrow 114 - 19\alpha = -3 \Rightarrow 19\alpha = 117$.Thus,$(19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36$.

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 $(19 \alpha-6 \beta)^2=$

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