Consider ΔABC with A≡(a$$→$$);B≡(b$$→$$) and C≡(c$$→$$) . If b→⋅$$(a$$→$$+c$$→$$)=b$$→⋅b→$$+a$$→⋅c→;|b→−a→|$$=3$$;|c→−b→|$$=4$$ then the angle between the medians AM→ and BD→ is :

Question

Consider ΔABC with A≡(a$$→$$);B≡(b$$→$$) and C≡(c$$→$$) . If b→⋅$$(a$$→$$+c$$→$$)=b$$→⋅b→$$+a$$→⋅c→;|b→−a→|$$=3$$;|c→−b→|$$=4$$  then  the angle between the medians AM→ and BD→ is :

Infinity Learn · Accepted Answer

b→ ⋅ a→ $$+$$ b→ ⋅ c→  $$=$$  b→ ⋅ b→  $$+$$  a→ ⋅ c→ or  b→⋅(a$$→−b→$$)−c→⋅(a$$→−b→$$)=0$$ or  $$(b$$→−c→$$)⋅$$(a$$→−b→$$)=0$$⇒BC and AB are perpendicular.  Now find angle between AM and BD $$M=4i^2=2i$$ $$^$$   $$AM=2i^-3j^D=4i^+3j^2$$   $$BD=4i^+3j^2If$$ θ is the angle between AM and BD then  $$cos$$θ$$=AM$$·$$BDAMBD=8-9213254=-1513$$

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

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Consider ΔABC with A≡(a→);B≡(b→) and C≡(c→) . If b→⋅(a→+c→)=b→⋅b→+a→⋅c→;|b→−a→|=3;|c→−b→|=4 then the angle between the medians AM→ and BD→ is :

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