The two vectors hat{i}+hat{j}+hat{k} and hat{ | vedclass

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(D) Let $\vec{AB} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{AC} = \hat{i}+3\hat{j}+5\hat{k}$.Let $M$ be the midpoint of side $BC$. The median through $A$ i

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$\sqrt{14}$

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(D) Let $\vec{AB} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{AC} = \hat{i}+3\hat{j}+5\hat{k}$.Let $M$ be the midpoint of side $BC$. The median through $A$ is represented by the vector $\vec{AM}$.In $	riangle ABC$,by the triangle law of vector addition,the position vector of the midpoint $M$ of $BC$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:$\vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC})$Substituting the given vectors:$\vec{AM} = \frac{1}{2}((\hat{i}+\hat{j}+\hat{k}) + (\hat{i}+3\hat{j}+5\hat{k}))$$\vec{AM} = \frac{1}{2}(2\hat{i} + 4\hat{j} + 6\hat{k})$$\vec{AM} = \hat{i} + 2\hat{j} + 3\hat{k}$The length of the median is the magnitude of vector $\vec{AM}$:$|\vec{AM}| = \sqrt{(1)^2 + (2)^2 + (3)^2}$$|\vec{AM}| = \sqrt{1 + 4 + 9}$$|\vec{AM}| = \sqrt{14}$

The two vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}+5\hat{k}$ represent the two sides $\vec{AB}$ and $\vec{AC}$ respectively of a $\triangle ABC$. The length of the median through $A$ is

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