The two vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ represent the two sides $\overline{A B}$ and $\overline{A C}$ respectively of a $\triangle A B C$. The length of the median through $A$ is

• A

  $\frac{\sqrt{14}}{2}$
• B

  14
• C

  7
• Correct answer
  D

  $\sqrt{14}$

Solution

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

$\begin{array}{ll}\therefore & \mathbf{A B}+\mathbf{B C}+\mathbf{C A}=0 \\ \Rightarrow \quad & \mathbf{B C}=\mathbf{A C}-\mathbf{A B} \\ \Rightarrow \mathbf{B} \mathbf{M}= & \frac{\mathbf{A C}-\mathbf{A B}}{2}(\text { since, } M \text { is a mid-point of } B C)\end{array}$ And also, $\mathbf{A B}+\mathbf{B} \mathbf{M}+\mathbf{M A}=0$ $\Rightarrow \quad \mathbf{A B}+\frac{\mathbf{A C}-\mathbf{A B}}{2}=\mathbf{A M}$ $\Rightarrow \quad \mathbf{A M}=\frac{\mathbf{A B}+\mathbf{A C}}{2}$ $\Rightarrow \quad \mathbf{A M}=\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{2}$ $\Rightarrow \quad \mathbf{A M}=\frac{2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{2}$ $\Rightarrow \quad \mathbf{A M}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ $\therefore \quad|\mathbf{A M}|=\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}$ $=\sqrt{1+4+9}=\sqrt{14}$

Hence, the correct answer is option: 'D'.