Question

The vector $\overline{AB}=3\hat{i}+4\hat{j}$ and $\overline{AC}=5\hat{i}-2\hat{j}+4\hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

• A. $\sqrt{21}$
• B. $\frac{6}{\sqrt{3}}$
• C. $\frac{33}{\sqrt{2}}$
• D. $\sqrt{33}$

Answer 1

The correct option is A $\sqrt{21}$

We know that the sum of three vectors of a triangle is $zero$.

$\therefore AB+BC+CA=0$

or, $BC=AC-AB$ [ $\therefore AC=-CA$]

$BD=\frac{AC-AB}{2}\left[D\right]$ the mid point of $BC$]

Also,$AB+BD+DA=0$ [Properties of a triangle]

or , $\begin{array}{c}\Rightarrow AB+\frac{AC-AB}{2}=AD\left[AD=-DA\right]\\ \Rightarrow AD=\frac{AB+AC}{2}=\frac{3\hat{i}+4\hat{j}+5\hat{i}-2\hat{j}+4\hat{k}}{2}\\ =\frac{8\hat{i}+2\hat{j}+4\hat{k}}{2}=4\hat{i}+\hat{j}+2\hat{k}\\ \therefore \left|AD\right|=\sqrt{4^2+1^2+2^2}=\sqrt{16+1+4}\\ =\sqrt{21}\end{array}$

Then,

Option $A$ is correct answer.