Question

The $2$ vectors $\hat{j}+\hat{k}$ and $3\hat{i}-\hat{j}+4\hat{k}$ represents the two sides $AB$ and $AC$, respectively of a $△ABC$. The length of the median through $A$ is

• A. $\frac{\sqrt{34}}{2}$
• B. $\frac{\sqrt{48}}{2}$
• C. $\sqrt{18}$
• D. $Noneofthese$

Answer 1

Answer: (A)

$\frac{\sqrt{34}}{2}$

Given that,

$\begin{array}{c}AB=\hat{j}+\hat{k}\\ AC=3\hat{i}-\hat{j}+4\hat{k}\end{array}$

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

$\therefore AB+BC+CA=0$

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

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

Also,

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

or, $AD=\frac{AB+AC}{2}$

$\begin{array}{c}=\frac{\hat{j}+\hat{k}+3\hat{i}-\hat{j}+4\hat{k}}{2}\\ =\frac{3\hat{i}+5\hat{k}}{2}=\frac{3\hat{i}}{2}+\frac{5\hat{i}}{2}\\ \therefore \left|AD\right|=\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{5}{2}\right)}^2}\\ =\sqrt{\frac{9}{4}+\frac{25}{4}}\\ =\sqrt{\frac{34}{4}}=\frac{\sqrt{34}}{2}\end{array}$

Hence,

Option $A$ is correct answer.