Let A, B, C be three points whose position ve | vedclass

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(A) The points $A, B, C$ are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.$\overrightarrow{AB} = \vec{b} - \v

Vedclass · Accepted Answer

$\frac{\sqrt{82}}{2}$

Vedclass · Answer

(A) The points $A, B, C$ are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (\alpha-4)\hat{j} + (4-3)\hat{k} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$.$\overrightarrow{AC} = \vec{c} - \vec{a} = (3-1)\hat{i} + (-2-4)\hat{j} + (5-3)\hat{k} = 2\hat{i} - 6\hat{j} + 2\hat{k}$.For collinearity,the ratios of components must be equal: $\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$.Solving $\frac{\alpha-4}{-6} = \frac{1}{2}$ gives $\alpha-4 = -3$,so $\alpha = 1$.Since the points are non-collinear for $\alpha 
eq 1$,the smallest positive integer $\alpha$ for which they are non-collinear is $\alpha = 2$.Now,find the midpoint $M$ of $BC$: $M = \frac{\vec{b} + \vec{c}}{2} = \frac{(2+3)\hat{i} + (2-2)\hat{j} + (4+5)\hat{k}}{2} = \frac{5}{2}\hat{i} + 0\hat{j} + \frac{9}{2}\hat{k}$.The length of the median $AM$ is the magnitude of $\vec{M} - \vec{A} = (\frac{5}{2}-1)\hat{i} + (0-4)\hat{j} + (\frac{9}{2}-3)\hat{k} = \frac{3}{2}\hat{i} - 4\hat{j} + \frac{3}{2}\hat{k}$.$AM = \sqrt{(\frac{3}{2})^2 + (-4)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{18}{4} + 16} = \sqrt{4.5 + 16} = \sqrt{20.5} = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{2}$.

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