If a , b and c are unit vectors satisfying | a - b |^ 2 +| b - c |^ 2 +| c - a |^ 2 =9 , then |2 a +5 b +5 c | is

Correct answer: 3 array l | a |=| b |=| c |=1 | a - b |^ 2 +| b - c |^ 2 +| c - a |^ 2 =9 2(| a |^ 2 +| b |^ 2 +| c |^ 2 )-2( a · b + b · c + c · a )=9 a · b + b · c + c · a =(-3)/(2) array Also, array l | a + b + c |^ 2 =| a |^ 2 +| b |^ 2 +| c |^ 2 +2( a · b + b · c + c · a ) =1+1+1+2 ×(-(3)/(2))=0 a + b + c =0 ( b + c )=- a |2 a +5( b + c )|=|2 a -5 a |=|-3 a |=3 array

JEE Advanced 2012Paper 1

Exam: JEE Advanced
Year: 2012
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2012 · Paper 1 · Mathematics · Vector Algebra

If a , b and c are unit vectors satisfying | a - b |^ 2 +| b - c |^...

\vec{a}, \vec{b}

and

\vec{c}

are unit vectors satisfying

|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9

, then

|2 \vec{a}+5 \vec{b}+5 \vec{c}|

Correct answer

Step-by-step solution

\begin{array}{l} \because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\ |\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\ \Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\ \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2} \end{array}

Also,

\begin{array}{l} |\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ =1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\ \Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\ \therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3 \end{array}

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