A,B,C have position vectors a= i+4 j+3 k, b=2 i+ j+4 k, c=3 i-2 j+5 k.

Question

$A,B,C$ have position vectors $\vec a=\hat i+4\hat j+3\hat k$, $\vec b=2\hat i+\alpha\hat j+4\hat k$, $\vec c=3\hat i-2\hat j+5\hat k$. If $\alpha$ is the smallest positive integer for which $A,B,C$ are non-collinear, the length of the median through $A$ in $	riangle ABC$ is:

Accepted Answer

Collinearity of $\vec{AB},\vec{AC}$ requires their components proportional; checking gives $\alpha=1$ as the collinear case, so the smallest positive integer making them non-collinear is $\alpha=2$.

With $B=(2,2,4)$, $C=(3,-2,5)$: midpoint of $BC$ is $M=\left(\dfrac52,0,\dfrac92\right)$.

$$\vec{AM} = \left(\frac52-1,\ 0-4,\ \frac92-3\right) = \left(\frac32,-4,\frac32\right)$$
$$|AM| = \sqrt{\frac94+16+\frac94} = \sqrt{\frac{82}4} = \frac{\sqrt{82}}2$$

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