The vectors overrightarrow{AB} = 3hat{i} + 5h | vedclass

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(C) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.Since $D$ is the midpoint of $BC$,the position ve

Vedclass · Accepted Answer

$5$

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(C) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.Since $D$ is the midpoint of $BC$,the position vector of $D$ relative to $A$ is given by $\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$.Substituting the given vectors:$\overrightarrow{AD} = \frac{1}{2}((3\hat{i} + 5\hat{j} + 4\hat{k}) + (5\hat{i} - 5\hat{j} + 2\hat{k}))$$\overrightarrow{AD} = \frac{1}{2}(8\hat{i} + 0\hat{j} + 6\hat{k})$$\overrightarrow{AD} = 4\hat{i} + 3\hat{k}$The length of the median is the magnitude of $\overrightarrow{AD}$:$|\overrightarrow{AD}| = \sqrt{4^2 + 0^2 + 3^2}$$|\overrightarrow{AD}| = \sqrt{16 + 9} = \sqrt{25} = 5 	ext{ units}$.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

The vectors $\overrightarrow{AB} = 3\hat{i} + 5\hat{j} + 4\hat{k}$ 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 ............. $unit$.

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