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Question: For non – zero vectors \(\vec{a}\) and \(\vec{b}\) if \(|\vec{a} + \vec{b} | < |\vec{a} - \vec{b} | ...

For non – zero vectors a\vec{a} and b\vec{b} if a+b<ab|\vec{a} + \vec{b} | < |\vec{a} - \vec{b} | , then a\vec{a} and b\vec{b} are
a) Collinear
b) Perpendicular to each other
c) Inclined at an acute angle
d) Inclined at an obtuse angle

Explanation

Solution

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: a\vec{a} and b\vec{b} are non – zero vectors.
a+b<ab|\vec{a} + \vec{b} | < |\vec{a}- \vec{b} | .
Squaring both sides.
a2+b2+2ab cosθ<a2+b22ab cosθa^{2} + b^{2} + 2ab\ cos \theta < a^{2} + b^{2} - 2ab\ cos \theta
2ab cosθ<2ab cosθ2ab\ cos \theta <-2ab\ cos \theta
4ab cosθ<04ab\ cos \theta < 0
θ\theta is the angle between a\vec{a} and b\vec{b}.
a\vec{a} and b\vec{b} are non – zero vectors.
a\vec{a} and b\vec{b} are Positive.
a,b>0a, b >0
cosθ<0\therefore cos\theta <0
Cosine function is negative in the second and third quadrant.
So, a\vec{a} and b\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 9090^{\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.