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CBSE Class 12 Math 2012 Solved Paper

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Question : 11 of 29
Marks: +1, -0
If a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} are three vectors such that ∣a⃗∣|\vec{a}| = 5 , ∣b⃗∣|\vec{b}| = 12 and ∣c⃗∣|\vec{c}| = 13 and a⃗+b⃗+c⃗\vec{a}+\vec{b}+\vec{c} = 0. Find the value of a⃗⋅b⃗\vec{a}\cdot\vec{b} + b⃗⋅c⃗\vec{b}\cdot\vec{c} + c⃗⋅a⃗\vec{c}\cdot\vec{a}
Solution:  
Considering dot product on both sides,
(a⃗+b⃗+c⃗)(\vec{a}+\vec{b}+\vec{c}) . (a⃗+b⃗+c⃗)(\vec{a}+\vec{b}+\vec{c}) = 0 . 0
⇒ ∣a⃗∣2+∣b⃗∣2+∣c⃗∣2|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2 (a⃗⋅b⃗(\vec{a}\cdot\vec{b} + b⃗⋅c⃗\vec{b}\cdot\vec{c} + c⃗⋅a⃗)\vec{c}\cdot\vec{a}) = 0
⇒ 52+122+1325^2+12^2+13^2 + 2 (a⃗⋅b⃗(\vec{a}\cdot\vec{b} + b⃗⋅c⃗\vec{b}\cdot\vec{c} + c⃗⋅a⃗)\vec{c}\cdot\vec{a}) = 0 = 0
⇒ 2 (a⃗⋅b⃗(\vec{a}\cdot\vec{b} + b⃗⋅c⃗\vec{b}\cdot\vec{c} + c⃗⋅a⃗)\vec{c}\cdot\vec{a}) = - 388
⇒ (a⃗⋅b⃗(\vec{a}\cdot\vec{b} + b⃗⋅c⃗\vec{b}\cdot\vec{c} + c⃗⋅a⃗)\vec{c}\cdot\vec{a}) = - 3382\frac{338}{2} = - 169
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