JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 9)
Let A, B, C be three points whose position vectors respectively are
$$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$$
$$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$$
$$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$$
If $$\alpha$$ is the smallest positive integer for which $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are noncollinear, then the length of the median, in $$\Delta$$ABC, through A is :
$${{\sqrt {82} } \over 2}$$
$${{\sqrt {62} } \over 2}$$
$${{\sqrt {69} } \over 2}$$
$${{\sqrt {66} } \over 2}$$
Explanation
$\overrightarrow{A B} \| \overrightarrow{A C}$ 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 $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$
$A M=\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$
$\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2}$
$\Rightarrow \alpha=1$
$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)
Mid point of $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$
$A M=\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$
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