Let A , B , C be three points whose position vectors respectively are: a = hat i +4 hat j +3 hat k b =2 hat i +α hat j +4 hat k , α ∈ R c =3 hat i -2 hat j +5 hat k If α is the smallest positive integer for which veca, vecb, vecc are non-collinear, then the length of the median, in triangle ABC, through A is:

Question

Let A , B , C be three points whose position vectors respectively are: a = hat i +4 hat j +3 hat k b =2 hat i +α hat j +4 hat k , α ∈ R c =3 hat i -2 hat j +5 hat k If α is the smallest positive integer for which veca, vecb, vecc are non-collinear, then the length of the median, in triangle ABC, through A is: (A) (√82/2) (B) (√62/2) (C) (√69/2) (D) (√66/2)

Tardigrade Admin · Accepted Answer

AB || AC if (1/2)=(α-4/-6)=(1/2) ⇒ α=1 veca , vecb , vecc are non-collinear for α=2 (smallest positive integer) Mid-point of BC = M ((5/2), 0, (9/2)) AM =√(9/4)+16+(9/4)=(√82/2)

$\frac{82}{2}$

$\frac{62}{2}$

$\frac{69}{2}$

$\frac{66}{2}$

$A B ∣∣ A C$ if $\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2} \Rightarrow α = 1$
$a, b, c$ are non-collinear for $α = 2$ (smallest positive integer)
Mid-point of $BC = M (\frac{5}{2}, 0, \frac{9}{2})$
$A M = \frac{9}{4} + 16 + \frac{9}{4} = \frac{82}{2}$

Q. Let A,B,C be three points whose position vectors respectively are: a=i^+4j^​+3k^ b=2i^+αj^​+4k^,α∈R c=3i^−2j^​+5k^ If α is the smallest positive integer for which a,b,c are non-collinear, then the length of the median, in △ABC, through A is:

282​​

262​​

269​​

266​​

AB∣∣AC if 21​=−6α−4​=21​⇒α=1 a,b,c are non-collinear for α=2 (smallest positive integer) Mid-point of BC=M(25​,0,29​) AM=49​+16+49​​=282​​

$\frac{82}{2}$

$\frac{62}{2}$

$\frac{69}{2}$

$\frac{66}{2}$

$A B ∣∣ A C$ if $\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2} \Rightarrow α = 1$
$a, b, c$ are non-collinear for $α = 2$ (smallest positive integer)
Mid-point of $BC = M (\frac{5}{2}, 0, \frac{9}{2})$
$A M = \frac{9}{4} + 16 + \frac{9}{4} = \frac{82}{2}$