If $\overline{a}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}$,$\overline{b}=4 \hat{\imath}+5 \hat{\jmath}+3 \hat{k}$ and $\overline{c}=6 \hat{\imath}+\hat{\jmath}+5 \hat{k}$ are the position vectors of the vertices of a triangle $ABC$ respectively,then the position vector of the intersection of the medians (centroid) of the triangle $ABC$ is:

  • A
    $4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • B
    $2 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • C
    $5 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • D
    $3 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}$

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In a regular hexagon $ABCDEF$,$\overrightarrow{AB}=\vec{a}$ and $\overrightarrow{BC}=\vec{b}$,then $\overrightarrow{FA}=$

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:

If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

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