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JEE Mains · Maths · STD 12 - 10. vector algebra

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

  1. A \(\frac{\sqrt{82}}{2}\)
  2. B \(\frac{\sqrt{62}}{2}\)
  3. C \(\frac{\sqrt{69}}{2}\)
  4. D \(\frac{\sqrt{66}}{2}\)
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Answer & Solution

Correct Answer

(A) \(\frac{\sqrt{82}}{2}\)

Step-by-step Solution

Detailed explanation

\(\overline{ AB } \| \overline{ AC }\) 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 \(BC = M \left(\frac{5}{2}, 0, \frac{9}{2}\right)\)…
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