JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q15

← JEE Main 2022 · 29 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 15 of 90

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 :

\dfrac{\sqrt{82}}2

\dfrac{\sqrt{62}}2

\dfrac{\sqrt{69}}2

\dfrac{\sqrt{66}}2

Answer and worked solution

Official answer: A

Collinear when $(1,\alpha-4,1)\parallel(2,-6,2)$ , i.e. $\alpha=1$ . So $\alpha=2$ .

Median. Midpoint of BC is $\left(\frac52,0,\frac92\right)$ ; from A: $\left(\frac32,-4,\frac32\right)$ , length $\frac{\sqrt{82}}2$ .

(A)

Topics:Vector operations

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JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 15