Q101 / 155 · WBJEE 2016 · Mathematics

For non-zero verctors a and b, $\mid a+b \mid < \mid a-b\mid$, then a and b are

  • A
    collinear
  • B
    perpendicular to each other
  • C
    inclined at an acute angle
  • Correct answer
    D
    inclined at an obtuse angle

Solution

Given, $\quad|\mathrm{a}+\mathrm{b}| < |\mathrm{a}-\mathrm{b}|$ Now, $|\mathbf{a}+\mathbf{b}|^{2} < |\mathbf{a}-\mathbf{b}|^{2}$ $\Rightarrow \quad(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}+\mathbf{b}) < (\mathbf{a}-\mathbf{b})(\mathbf{a}-\mathbf{b})$ $\Rightarrow \quad|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b} < |\mathbf{a}|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$ $\Rightarrow \quad 4 \mathrm{a} \cdot \mathrm{b} < 0$ $\Rightarrow \quad \mathbf{a} \cdot \mathbf{b} < 0$ $\Rightarrow \quad|\mathbf{a}||\mathbf{b}| \cos \theta < 0$ $\Rightarrow \quad \cos \theta < 0$ $\Rightarrow \quad \theta \in$ obluse angle $\therefore ~a$ and $b$ are inclined at an obtuse angle.

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

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