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Let $\mathrm{A,B,C}$ be three points whose position vectors respectively are

$\begin{aligned} &\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k} \\ &\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in \mathbb{R} \\ &\vec{c}=3 \hat{i}-2 \hat{j}+5 \hat{k} \end{aligned}$

If $\mathrm{\alpha }$ is the smallest positive integer for which $\mathrm{ } \vec{a}$ , $\mathrm{ } \vec{b}$ , $\mathrm{ } \vec{c}$ are non-collinear, then the length of the median in $\mathrm{\Delta ABC }$ , through $\mathrm{A }$ is:
Option: 1
$\frac{\sqrt{82}}{2}$

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

Option: 3
$\frac{\sqrt{69}}{2}$

Option: 4
$\frac{\sqrt{66}}{2}$

Answers (1)

best_answer

$\mathrm{Let\: A(i+4 j+3 k), B(2 i+\alpha j+4 k) \: and \: C(3 i-2 j+5 k)}$

If $\mathrm{A,B\: and \: C}$ are collinear, then

$\mathrm{ \overrightarrow{A C}=\lambda(\overrightarrow{B C}) \quad \text { for any } \lambda \in R} \\$

$\mathrm{\Rightarrow (2 i-6 j+2 k)=\lambda(i+(-2-\alpha) j+k)} \\$

$\mathrm{\Rightarrow 2=\lambda,-6=\lambda(-2-\alpha), \quad 2=\lambda} \\$

$\mathrm{\Rightarrow -6=2(-2-\alpha) }$

$\mathrm{\Rightarrow -3=-2-\alpha} \\$

$\Rightarrow \alpha = 1$

$\therefore$ For $\mathrm{\alpha \neq 1, A, B,C}$ are non-collinear

Smallest positive integral $\mathrm{\alpha =2}$

$\mathrm{D}$ using mid-point formula is

$\mathrm{\left(\frac{5}{2}, 0, \frac{9}{2}\right)} \\$

$\mathrm{A D =\sqrt{\left(\frac{5}{2}-1\right)^{2}+(0-4)^{2}+\left(\frac{9}{2}-3\right)^{2}}} \\$

$\mathrm{=\sqrt{\frac{9}{4}+16+\frac{9}{4}}} \\$

$\mathrm{=\sqrt{\frac{9+64+9}{4}} }$

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

Hence answer is option 1

Posted by

himanshu.meshram

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