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