Let $A, B, C$ be three points whose position vectors respectively are $\bar{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$ , $\bar{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}$ , $α \in R$ , $\bar{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$ . If $α$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, $Δ A B C$ through A is

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$\frac{\sqrt{66}}{2}$

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

$\frac{\sqrt{69}}{2}$

$\frac{\sqrt{62}}{2}$

answer is A.

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Detailed Solution

Let us assume that vectors are collinear
$\vec{A B} | | \vec{A C}$ if $\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2} \Rightarrow α = 1$
Now, $\bar{a}, \bar{b}, \bar{c}$ to be non-collinear
Smallest positive integer will be $α = 2$
So, mid point of $B C = m (\frac{5}{2}, 0, \frac{9}{2})$
$∴ A M = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \frac{\sqrt{82}}{2}$