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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
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
  • ✓
    $\frac{\sqrt{82}}{2}$
  • B
    $\frac{\sqrt{62}}{2}$
  • C
    $\frac{\sqrt{69}}{2}$
  • D
    $\frac{\sqrt{66}}{2}$
✓

Answer

Correct option: A.
$\frac{\sqrt{82}}{2}$
a
$\overline{ AB } \| \overline{ AC }$ 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 $BC = M \left(\frac{5}{2}, 0, \frac{9}{2}\right)$

$AM =\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$

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