Q43 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Mathematics

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 ........

  • A
    \(30^{\circ}\)
  • B
    \(120^{\circ}\)
  • C
    \(90^{\circ}\)
  • Correct answer
    D
    \(60^{\circ}\)

Solution

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, \(\begin{aligned} \Rightarrow & 2 \mathbf{a} \cdot \mathbf{b} =49-9-25 \\ \Rightarrow & 2 a \cdot b \cos \theta =15 \\ \Rightarrow & \cos \theta =\frac{15}{2 \times 3 \times 5} \\ & =\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ} \end{aligned}\)

Hence, the correct answer is option: 'D'.

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