Question

For non-zero vectors$a$ and $b,$ if $|\mathrm{a}+\mathrm{b}|<|\mathrm{a}-\mathrm{b}|,$ then $a$ and $b$ are:

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

Answer 1

The correct option is D

inclined at an obtuse angle

Step 1. Simplify $\left|a+b\right|<\left|a-b\right|$

Let $\theta$ be the angle between the vectors $\text{'}\mathrm{a}\text{'}\mathrm{and}\text{'}\mathrm{b}\text{'}$

Given,

For non-zero vectors $\text{'}a\text{'}$ and $\text{'}b\text{'}$

$\left|a+b\right|<\left|a-b\right|$

$\therefore$$\Rightarrow a^2+b^2+2ab\cos \theta <a^2+b^2-2ab\cos \theta$

$\Rightarrow 4ab\cos \theta <0$

$\therefore \cos \theta <0$ $\left[\because 4>0;a,b>0\right]$

Step2. Calculate the value of $\theta$

$\cos \theta$is negative in the second and third quadrants.

$\therefore \theta \in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)$

Thus, $\text{'}a\text{'}$ and $\text{'}b\text{'}$ inclined at an obtuse angle.

Hence, Option(D) is the correct answer.