First slide

Dot product or scalar product of vectors

Question

$Given a parallelogram ABCD . If | \vec{A B ∣} = a, | \vec{A D} ∣= b and | \vec{A C} | = c, then \vec{D B} \cdot \vec{A B}$ $has the value :$

Easy

A

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

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

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

$n o n e o f t h e s e$

Solution

$To find (\vec{a} - \vec{b}) \cdot \vec{a}, i.e. | \vec{a} |^{2} - \vec{a} \cdot \vec{b}$ ………..(1)

$\begin{aligned} Now \\ \vec{a} + \vec{b} = \vec{c} \end{aligned}$
$\Rightarrow | \vec{a} |^{2} + | \vec{b} |^{2} + 2 \vec{a} \cdot \vec{b} = | \vec{c} |^{2}$
$\Rightarrow 2 \vec{a} \cdot \vec{b} = | \vec{c} |^{2} - | \vec{a} |^{2} - | \vec{b} |^{2} \vec{a} \cdot \vec{b} = \frac{| \vec{c} |^{2} - | \vec{a} |^{2} - | \vec{b} |^{2}}{2}$
$Substitute the value of \vec{a} \cdot \vec{b} in eqn. (1)$
$|\vec{a^{2}}| -$ $\frac{| \vec{c} |^{2} - | \vec{a} |^{2} - | \vec{b} |^{2}}{2}$
$\frac{3 {\vec{a}}^{2} + {\vec{b}}^{2} - {\vec{c}}^{2}}{2}$

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