For a triangle ABC,if vec{AB} = 3hat{i} + 4ha | vedclass

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(D) Let $D$ be the midpoint of side $BC$. The median from vertex $A$ is the vector $\vec{AD}$.Since $D$ is the midpoint of $BC$,$\vec{AD} = \frac{1}{2

Vedclass · Accepted Answer

$\sqrt{33}$

Vedclass · Answer

(D) Let $D$ be the midpoint of side $BC$. The median from vertex $A$ is the vector $\vec{AD}$.Since $D$ is the midpoint of $BC$,$\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$.Given $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$.$\vec{AD} = \frac{1}{2}((3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$.$\vec{AD} = \frac{1}{2}(8\hat{i} - 2\hat{j} + 8\hat{k}) = 4\hat{i} - \hat{j} + 4\hat{k}$.The length of the median $AD$ is the magnitude of vector $\vec{AD}$.$|\vec{AD}| = \sqrt{4^2 + (-1)^2 + 4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}$.

For a triangle $ABC$,if $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$,then the length of the median drawn from vertex $A$ is:

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