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$3\,\,\overrightarrow{OD}+\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}=$
$\mathbf{p}=2\mathbf{a}-3\mathbf{b},\,\,\,\mathbf{q}=\mathbf{a}-2\mathbf{b}+\mathbf{c},\,\,\mathbf{r}=-3\mathbf{a}+\mathbf{b}+2\mathbf{c};$ where a, b and c being non-zero, non-coplanar vectors, then the vector $-2\mathbf{a}+3\mathbf{b}-\mathbf{c}$ is equal to
In a trapezium, the vector $\overrightarrow{BC}=\lambda \overrightarrow{AD}.$ We will then find that $\mathbf{p}=\overrightarrow{AC}+\overrightarrow{BD}$ is collinear with $\overrightarrow{AD},$ If $\mathbf{p}=\mu \overrightarrow{AD},$ then
If $\mathbf{a}=2\mathbf{i}+\mathbf{j}-8\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+3\mathbf{j}-4\mathbf{k},$ then the magnitude of $\mathbf{a}+\mathbf{b}=$
A, B, C, D, E are five coplanar points, then $\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}+\overrightarrow{AE}+\overrightarrow{BE}+\overrightarrow{CE}$ is equal to
If $\mathbf{a}=3\mathbf{i}-2\mathbf{j}+\mathbf{k},\,\,\mathbf{b}=2\mathbf{i}-4\mathbf{j}-3\mathbf{k}$ and $\mathbf{c}=-\mathbf{i}+2\mathbf{j}+2\mathbf{k},$ then $\mathbf{a}+\mathbf{b}+\mathbf{c}$ is
Five points given by A, B, C, D, E are in a plane. Three forces $\overrightarrow{AC},\,\,\overrightarrow{AD}$ and $\overrightarrow{AE}$ act at A and three forces $\overrightarrow{CB},\,\,\overrightarrow{DB},\,\,\overrightarrow{EB}$ act at B. Then their resultant is
The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12N. The magnitude of the two forces are
The unit vector parallel to the resultant vector of $2\mathbf{i}+4\mathbf{j}-5\mathbf{k}$ and $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ is
If a, b, c are the position vectors of the vertices A, B, C of the triangle ABC, then the centroid of $\Delta \,ABC$ is

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