JEE Main 2022 (29 Jun, Shift 2) Maths Q15: Vector Algebra

Question

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

Accepted Answer

Option 1: $\frac{\sqrt{82}}{2}$

Answer

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

Answer

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

Answer

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

Answer

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

JEE Main 29 June 2022, Shift 2, Maths Q15

Same idea in other shifts