Question

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

• A. $\sqrt{13}\mathrm{units}$
• B. $2\sqrt{5}\mathrm{units}$
• C. $5\mathrm{units}$
• D. $10\mathrm{units}$

Answer 1

The correct option is C

$5\mathrm{units}$

Explanation for correct option:

Given, $\overrightarrow{AB}=3\hat{i}+5\hat{j}+4\hat{k}$ and $\overrightarrow{AC}=5\hat{i}-5\hat{j}+2\hat{k}$

Let $AD$ be the median of the triangle

Position vector $\overrightarrow{AD}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)$
$\overrightarrow{AD}=\frac{1}{2}\left[\left(3+5\right)\hat{i}+\left(5-5\right)\hat{j}+\left(4+2\right)\hat{k}\right]$
$$\begin{gathered} \overrightarrow{AD}=\frac{1}{2}\left(8\hat{i}+6\hat{k}\right) \\ =4\hat{i}+3\hat{k} \end{gathered}$$

Length of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\sqrt{{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}^{\mathbf{2}}}$

Thus, Length of the median $\left|\overrightarrow{AD}\right|=\sqrt{4^2+3^2}=\sqrt{25}=5$

Hence, option C is correct.