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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)

best_answer

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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