Question

Resultant of two vectors is perpendicular to smaller one & it's magnitude is root3/2 times bigger one.
Then angle between them is
(a) 60 (b) 120
(c) 30 (d) 150

Answer 1

Answer :

The angel between the two vectors is 120°

Given :

• Resultant of two vectors is perpendicular to the vector which is smaller
• The magnitude of the resultant vector is √3/2 times the bigger vector.

To Find :

• The angle between the two vectors

Formula to be used :

If θ is the angle between two vectors A and B then the magnitude of it's resultant is given by :

$\bullet \: \: \sf R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$

Dot product of two vectors A and B is given by

$\sf \bullet \: \: \vec{A}.\vec{B} = AB\cos\theta$

Solution :

Let the vectors be A and B and the resultant be R . The smaller vector be B and bigger be A . Also let the angle between the vectors A and B be θ

$\sf R = \dfrac{\sqrt{3}}{2}A \\\\ \sf \implies R^{2} = \dfrac{3}{4}A \\\\ \sf \implies A^{2} + B^{2} + 2AB\cos\theta = \dfrac{3}{4}A^{2} \\\\ \sf \implies B^{2} + 2AB\cos\theta = \dfrac{-A^{2}}{4} ..........(1)$

Since the vectors R and B are perpendicular so from dot product of vector :

$\sf \vec{R}.\vec{B} = 0 \\\\ \sf \implies (\vec{A} + \vec{B} )\vec{B} = 0 \\\\ \sf \implies \vec{A}.\vec{B} + \vec{B}.\vec{B} = 0 \\\\ \sf \implies AB\cos\theta + B^{2} = 0 \\\\ \sf \implies B^{2} = -AB\cos\theta \\\\ \sf \implies B = -A\cos\theta \\\\ \implies \sf \cos\theta =- \dfrac{B}{A} ..........(2)$

Using the value of (2) in (1) :

$\sf \implies B^{2} + 2AB(-\dfrac{B}{A}) = - \dfrac{A^{2}}{4} \\\\ \sf \implies B^{2} - 2B^{2} = -\dfrac{A^{2}}{4} \\\\ \sf \implies B^{2} = \dfrac{A^{2}}{4} \\\\ \sf \implies B = \dfrac{A}{2} \\\\ \sf \implies A = 2B$

Now using the value of A in (2) :

$\sf \implies -\dfrac{B}{2B} = \cos\theta \\\\ \sf \implies \cos\theta = \dfrac{1}{2} \\\\ \sf \implies \cos\theta =- \cos60\degree \\\\ \sf \implies \cos\theta = \cos(180\degree - 60\degree) \\\\ \sf \implies \cos\theta = \cos120\degree \\\\ \sf \implies \theta = 120\degree$