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  4. Let A , B , C be three points whose position vectors respectively are: a = hat i +4 hat j +3 hat k b =2 hat i +α hat j +4 hat k , α ∈ R c =3 hat i -2 hat j +5 hat k If α is the smallest positive integer for which veca, vecb, vecc are non-collinear, then the length of the median, in triangle ABC, through A is:

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

JEE MainJEE Main 2022Vector Algebra

Solution:

$\overrightarrow{AB} || \overrightarrow{AC}$ if $\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2} \Rightarrow \alpha=1$
$\vec{a} , \vec{b} , \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)
Mid-point of $BC = M \left(\frac{5}{2}, 0, \frac{9}{2}\right)$
$AM =\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$