In    a    parallelogram ABCD ,   \left | \overrightarrow{AB} \right |=a,\: \left | \overrightarrow{AD} \right |=b\; and\: \left | \overrightarrow{AC} \right |=c,\; then\: \overrightarrow{DB}\cdot \overrightarrow{AB}\; has\: the\; value:

  • Option 1)

    \frac{1}{2}\left ( a^{2}-b^{2}+c^{2} \right )

  • Option 2)

    \frac{1}{4}\left ( a^{2}+b^{2}-c^{2} \right )

  • Option 3)

    \frac{1}{3}\left ( b^{2}+c^{2}-a^{2} \right )

  • Option 4)

    \frac{1}{2}\left ( a^{2}+b^{2}+c^{2} \right )

 

Answers (1)
P Plabita

As we learnt in

Scalar Product of two vectors (dot product) -

\vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta

- wherein

\Theta is the angle between the vectors\vec{a}\: and\:\vec{b}

Thus let A be position \vec{0} vector of B is \vec{a} , D is \vec{b}

\overrightarrow{DB}\ .\ \overrightarrow{AB}

=(\overrightarrow{B}-\overrightarrow{D}).(\overrightarrow{B}-\overrightarrow{A})

=\overrightarrow{B}. \overrightarrow{B}-\overrightarrow{B}.\overrightarrow{D}-\overrightarrow{A}.\overrightarrow{B}+\overrightarrow{A}.\overrightarrow{D}

=|\overrightarrow{b}|^{2}-\overrightarrow{a}.\overrightarrow{b}-0

=b^{2}-|\overrightarrow{a}|.|\overrightarrow{b}|cos\theta                                        ..............(i)

Also here,

\vec{a}+\vec{b}=\vec{c}

|\vec{a}+\vec{b}|=|\vec{c}|

a^{2}+b^{2}+2\vec{a}.\vec{b}=c^{2}

cos \theta =-\frac{(a^{2}+b^{2}-c^{2})}{2ab}

Put in (i)

\frac{b^{2}+a^{2}+b^{2}-c^{2}}{2}=\frac{3b^{2}+a^{2}-c^{2}}{2}

 

 


Option 1)

\frac{1}{2}\left ( a^{2}-b^{2}+c^{2} \right )

This option is correct.

Option 2)

\frac{1}{4}\left ( a^{2}+b^{2}-c^{2} \right )

This option is incorrect.

Option 3)

\frac{1}{3}\left ( b^{2}+c^{2}-a^{2} \right )

This option is incorrect.

Option 4)

\frac{1}{2}\left ( a^{2}+b^{2}+c^{2} \right )

This option is incorrect.

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