JEE Advanced · Mathematics · 30. Vector Algebra
If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are unit vectors satisfying \(|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9\), then \(|2 \vec{a}+5 \vec{b}+5 \vec{c}|\) is
- A 3
- B 8
- C 9
- D 1
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\
|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\
\Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\
\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}
\end{array}\)
Also,
\(\begin{array}{l}
|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\
=1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\
\Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\
\therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3
\end{array}\)
\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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