Q. The vectors $\overrightarrow{\mathbf{AB}} = 3\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}\ and\ \overrightarrow{\mathbf{AC}} = 5\widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

A) $\sqrt{13}$ units
B) $2\sqrt{5}$ units
C) 5 units
D) 10 units
Explanation

(c)
$\therefore Median\ vector\ through\ \overrightarrow{\mathbf{A}} = \frac{1}{2}\left( \overrightarrow{\mathbf{AB}} + \overrightarrow{\mathbf{AC}} \right)$
$= \frac{1}{2}\left\lbrack \left( 3\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}} \right) + \left( 5\widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) \right\rbrack$
$= 4\widehat{\dot{\mathbf{i}}} + 3\widehat{\mathbf{k}}$
$\therefore$ Length of the median $= \sqrt{4^{2} + 3^{2}} = 5$ units

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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