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Question: Answered & Verified by Expert
$\mathrm{If} \overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors inclined at an angle of $30^{\circ}$ to each other, then which one of the following is correct?
MathematicsVector AlgebraNDANDA 2007 (Phase 1)
Options:
  • A $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|>1$
  • B $\quad 1 < |\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| < 2$
  • C $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=2$
  • D $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|>2$
Solution:
2606 Upvotes Verified Answer
The correct answer is: $\quad 1 < |\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| < 2$
$\begin{aligned} &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{b}} \cos \theta \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=1+1+2|\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}| \cos 30^{\circ} \\ &=1+1+2 \times \frac{\sqrt{3}}{2} \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=2+\sqrt{3} \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{2+\sqrt{3}} \\ & 1 < \sqrt{2+\sqrt{3}} < 2 \\ \Rightarrow & 1 < |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| < 2 \end{aligned}$

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