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**Hint:**If a vector is non-zero, it has at least one non-zero component; a zero vector has all parts as zero, therefore, no length. A non-zero vector is one in which at least one non-zero is there, at least in absolute numbers. In general, a non-zero vector is not the identity element for the summation of the vector space.

**Complete step-by-step solution:**

Given: $\vec{a}$ and $\vec{b}$ are non – zero vectors.

$|\vec{a} + \vec{b} | < |\vec{a}- \vec{b} | $.

Squaring both sides.

$a^{2} + b^{2} + 2ab\ cos \theta < a^{2} + b^{2} - 2ab\ cos \theta$

$2ab\ cos \theta <-2ab\ cos \theta $

$4ab\ cos \theta < 0 $

$\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

$\vec{a}$ and $\vec{b}$ are non – zero vectors.

$\vec{a}$ and $\vec{b}$ are Positive.

$a, b >0$

$\therefore cos\theta <0$

Cosine function is negative in the second and third quadrant.

So, $\vec{a}$ and $\vec{b}$ are inclined at an obtuse angle.

**Option (d) is correct.**

**Note:**If the sum of two non-zero vectors is equal to their difference, then since the angle between given vectors is $90^{\circ}$, The vectors are perpendicular. The main difference between unit vector and non-zero vector is that the unit vector is the outcome of normalizing a non-zero vector and the unit vector is the ratio of vector to its length.

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