Question

The two vectors \[\hat{j} + \hat{k}\] and \[3 \hat{i} - \hat{j} + 4 \hat{k}\] represents the sides \[\overrightarrow{AB}\] and \[\overrightarrow{AC}\] respectively of a triangle ABC. Find the length of the median through A.

Answer 1

\[\overrightarrow{AB}\] as \[\hat{j} + \hat{k}\] In ∆ABC, \[\overrightarrow{AB} = \hat{j} + \hat{k}\] and \[\overrightarrow{AC} = 3 \hat{i} - \hat{j} + 4 \hat{k}\] 
Let the position vector of A be (0, 0, 0). Then, the position vectors of B and C are (0, 1, 1) and (3, −1, 4), respectively.

Suppose D be the mid-point of the line segment joining the points B(0, 1, 1) and C(3, −1, 4).
∴ Position vector of D \[= \frac{\left( \hat{j} + \hat{k} \right) + \left( 3 \hat{i} - \hat{j} + 4 \hat{k} \right)}{2} = \frac{3 \hat{i} + 5 \hat{k}}{2} = \frac{3}{2} \hat{i} + \frac{5}{2} \hat{k}\]
Now,
Length of the median, AD = \[\left| \overrightarrow{AD} \right| = \left| \left( \frac{3}{2} \hat{i} + \frac{5}{2} \hat{k} \right) - \left( 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \right) \right| = \left| \frac{3}{2} \hat{i} + \frac{5}{2} \hat{k} \right| = \sqrt{\left( \frac{3}{2} \right)^2 + 0^2 + \left( \frac{5}{2} \right)^2} = \sqrt{\frac{34}{4}} = \sqrt{\frac{17}{2}}\] units