First slide

Dot product or scalar product of vectors

Question

$Consider Δ A B C with A \equiv (\vec{a}); B \equiv (\vec{b}) and C \equiv (\vec{c}) . If$ $\vec{b} \cdot (\vec{a} + \vec{c}) = \vec{b} \cdot \vec{b} + \vec{a} \cdot \vec{c}; | \vec{b} - \vec{a} | = 3; | \vec{c} - \vec{b} | = 4 then$
$the angle between the medians$ $\vec{A M} and \vec{B D} is :$

Difficult

A

$π - \cos^{- 1} (\frac{1}{5 \sqrt{13}})$
B

$π - \cos^{- 1} (\frac{1}{13 \sqrt{5}})$
C

$\cos^{- 1} (\frac{1}{5 \sqrt{13}})$
D

$\cos^{- 1} (\frac{1}{13 \sqrt{5}})$

Solution

$\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = \vec{b} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
$or \vec{b} \cdot (\vec{a} - \vec{b}) - \vec{c} \cdot (\vec{a} - \vec{b}) = 0$
$or (\vec{b} - \vec{c}) \cdot (\vec{a} - \vec{b}) = 0$
$\Rightarrow BC and AB are perpendicular.$
$Now find angle between AM and BD$
$M = \frac{4 \hat{i}}{2} = 2 \hat{i} A M = 2 \hat{i} - 3 \hat{j}$
$D = \frac{4 \hat{i} + 3 \hat{j}}{2} B D = \frac{4 \hat{i} + 3 \hat{j}}{2}$
$I f θ i s t h e a n g l e b e t w e e n A M a n d B D t h e n \cos θ = \frac{A M \cdot B D}{|A M| |B D|}$
$= \frac{\frac{8 - 9}{2}}{\sqrt{13} \sqrt{\frac{25}{4}}} = - (\frac{1}{5 \sqrt{13}})$

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