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If \(\mathbf{a}+\mathbf{b}+\mathbf{c}=0\) and \(|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7\), then the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is ........
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Verified Answer
The correct answer is:
\(60^{\circ}\)
Given, \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}\)
\(\begin{array}{lllll}
\Rightarrow & (\mathbf{a}+\mathbf{b})=-\mathbf{c} \\
\Rightarrow & (\mathbf{a}+\mathbf{b})^2=\mathbf{c}^2 \\
\Rightarrow & \mathbf{a}^2+\mathbf{b}^2+2 \mathbf{a} \cdot \mathbf{b}=\mathbf{c}^2 \\
\Rightarrow & 2 \mathbf{a} \cdot \mathbf{b}=\mathbf{c}^2-\mathbf{a}^2-\mathbf{b}^2
\end{array}\)
As square of a vector \(=\) square of its magnitude,
\(\begin{aligned}
\Rightarrow & 2 \mathbf{a} \cdot \mathbf{b} =49-9-25 \\
\Rightarrow & 2 a \cdot b \cos \theta =15 \\
\Rightarrow & \cos \theta =\frac{15}{2 \times 3 \times 5} \\
& =\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
\end{aligned}\)
\(\begin{array}{lllll}
\Rightarrow & (\mathbf{a}+\mathbf{b})=-\mathbf{c} \\
\Rightarrow & (\mathbf{a}+\mathbf{b})^2=\mathbf{c}^2 \\
\Rightarrow & \mathbf{a}^2+\mathbf{b}^2+2 \mathbf{a} \cdot \mathbf{b}=\mathbf{c}^2 \\
\Rightarrow & 2 \mathbf{a} \cdot \mathbf{b}=\mathbf{c}^2-\mathbf{a}^2-\mathbf{b}^2
\end{array}\)
As square of a vector \(=\) square of its magnitude,
\(\begin{aligned}
\Rightarrow & 2 \mathbf{a} \cdot \mathbf{b} =49-9-25 \\
\Rightarrow & 2 a \cdot b \cos \theta =15 \\
\Rightarrow & \cos \theta =\frac{15}{2 \times 3 \times 5} \\
& =\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
\end{aligned}\)
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