question_answer

In a quadrilateral\[ABCD\], the point P divides \[DC\]in the ratio\[1:2\]and\[Q\]is the midpoint of\[AC\]. If\[\overrightarrow{\mathbf{AB}}+2\overrightarrow{\mathbf{AD}}+\overrightarrow{\mathbf{BC}}-2\overrightarrow{\mathbf{DC}}=k\overrightarrow{\mathbf{PQ}}\], then\[k\]is equal to

A)  -6                                          

B)  -4

C)  6                            

D)         4

Correct Answer: A

Solution :

Now,\[\overrightarrow{\mathbf{AB}}+2\overrightarrow{\mathbf{AD}}+\overrightarrow{\mathbf{BC}}-2\overrightarrow{\mathbf{DC}}\]                 \[=\overrightarrow{\mathbf{AC}}+2\overrightarrow{\mathbf{AD}}-2\overrightarrow{\mathbf{DC}}\]                 \[=\overrightarrow{\mathbf{AC}}+2(\overrightarrow{\mathbf{AC}}+\overrightarrow{\mathbf{CD}})-2\overrightarrow{\mathbf{DC}}\]                 \[=3\overrightarrow{\mathbf{AC}}-4\overrightarrow{\mathbf{DC}}\]                 \[=3(2\overrightarrow{\mathbf{QC}})-4\left( \frac{3}{2}\overrightarrow{\mathbf{PC}} \right)\]                 \[=6\overrightarrow{\mathbf{QC}}-6\overrightarrow{\mathbf{PC}}=6(\overrightarrow{\mathbf{QC}}+\overrightarrow{\mathbf{CP}})\] \[\Rightarrow \]                \[k\overrightarrow{\mathbf{PQ}}=6\overrightarrow{\mathbf{QP}}=6\overrightarrow{\mathbf{PQ}}\]                (given) \[\Rightarrow \]               \[k=-6\]