The vectors $\overrightarrow{\text{AB}} = 3\hat{\text{i}} + 4\hat{\text{k}}$ and $\overrightarrow{\text{AC}} = 5\hat{\text{i}} - 2\hat{\text{j}} + 4\hat{\text{k}}$ are the sides of a triangle ABC. The length of the median through A is
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To make vector calculations simpler, you can assign vertex A as the origin $(0,0,0)$. This makes the coordinates of point B $(3,0,4)$ and point C $(5,-2,4)$. Finding the midpoint of BC directly yields position D $(4,-1,4)$, and the distance formula gives $\sqrt{4^2+(-1)^2+4^2} = \sqrt{33}$!
Step 1: Picture the triangle and the median. We have a triangle ABC with vertex A as our reference point. The two sides leaving A are $\overrightarrow{AB}=3\hat{i}+4\hat{k}$ and $\overrightarrow{AC}=5\hat{i}-2\hat{j}+4\hat{k}$. A median from A runs to the midpoint of the opposite side BC, so let us call that midpoint D and find $\overrightarrow{AD}$. Step 2: Write the position-style relation for the midpoint. If we treat A as the origin, then B and C have position vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The midpoint D of segment BC has position vector equal to the average of B and C. Step 3: Form the median vector. Hence $\overrightarrow{AD}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$. This is just the average of the two side vectors. Step 4: Add the components. $\overrightarrow{AB}+\overrightarrow{AC}=(3+5)\hat{i}+(0-2)\hat{j}+(4+4)\hat{k}=8\hat{i}-2\hat{j}+8\hat{k}$. Step 5: Halve to get the median. $\overrightarrow{AD}=\dfrac{1}{2}(8\hat{i}-2\hat{j}+8\hat{k})=4\hat{i}-\hat{j}+4\hat{k}$. Step 6: Take the magnitude. $|\overrightarrow{AD}|=\sqrt{4^2+(-1)^2+4^2}=\sqrt{16+1+16}=\sqrt{33}$ units. So the length of the median through A is $\sqrt{33}$ units, matching option (1). \[ \boxed{\sqrt{33}\ \text{units}} \]