Points A(2,0,1) , B(1,-2,3) , C(0,4,2) are vertices of a…
Points $A(2,0,1)$, $B(1,-2,3)$, $C(0,4,2)$ are vertices of a triangle. Determine the median vector $\overrightarrow{AA_1}$ and its magnitude.
Explanation of Method
The median vector $\overrightarrow{AA_1}$ in triangle $ABC$.
Step 1: Midpoint $A_1$ of $BC$
\[ A_1 = \left(\frac{1+0}{2},\;\frac{-2+4}{2},\;\frac{3+2}{2}\right) = \left(\frac{1}{2},\;1,\;\frac{5}{2}\right) \]
Step 2: The median vector
\[ \overrightarrow{AA_1} = \left(\frac{1}{2}-2\right)\vec{i}+(1-0)\vec{j}+\left(\frac{5}{2}-1\right)\vec{k} \]
\[\overrightarrow{AA_1} = -\frac{3}{2}\,\vec{i}+\vec{j}+\frac{3}{2}\,\vec{k}\]
Step 3: Magnitude
\[ |\overrightarrow{AA_1}| = \sqrt{\frac{9}{4}+1+\frac{9}{4}} = \sqrt{\frac{22}{4}} = \frac{\sqrt{22}}{2} \]
\[|\overrightarrow{AA_1}| = \frac{\sqrt{22}}{2} \approx 2.345\]
- $\overrightarrow{AA_1}=-\tfrac32\vec i+\vec j+\tfrac32\vec k$
- $\text{sides of }\triangle ABC$
Step 4: Check via the median formula of Problem 19
\[ \overrightarrow{AB} = -\vec{i}-2\vec{j}+2\vec{k}, \qquad \overrightarrow{AC} = -2\vec{i}+4\vec{j}+\vec{k} \]
\[ \overrightarrow{AA_1} = \frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right) = \frac{1}{2}\left(-3\vec{i}+2\vec{j}+3\vec{k}\right) \;\checkmark \]
Step 5: Check via side lengths
\[ c = |AB| = \sqrt{1+4+4} = 3, \qquad b = |AC| = \sqrt{4+16+1} = \sqrt{21}, \qquad a = |BC| = \sqrt{1+36+1} = \sqrt{38} \]
\[ m_a = \frac{1}{2}\sqrt{2b^2+2c^2-a^2} = \frac{1}{2}\sqrt{42+18-38} = \frac{\sqrt{22}}{2} \;\checkmark \]
Result
\[\boxed{\,\overrightarrow{AA_1} = -\frac{3}{2}\,\vec{i}+\vec{j}+\frac{3}{2}\,\vec{k}\,}\]
\[\boxed{\,|\overrightarrow{AA_1}| = \frac{\sqrt{22}}{2} \approx 2.345\,}\]