Question

13.Resultant of two vectors is perpendicular to smaller one and it’s magnitude is root 3÷2 times bigger one. then angle between them is A.60° B.120° C. 30° D. 150

Answer 1

Given that,

Resultant of two vectors is perpendicular to smaller one. and it’s magnitude is $\dfrac{\sqrt{3}}{2}$ times bigger one.

Let the two vectors be a and b, and the angle between the two vectors be θ. Let the resultant of these two vectors be R, such that R is perpendicular to smaller vector a, and the the magnitude of R is $\dfrac{\sqrt{3}}{2}$ times of magnitude of bigger vector b.

$R=\dfrac{\sqrt{3}}{2}b$

From vector algebra,

We know that, the magnitude of the resultant vector R is given by

$R^2=a^2+b^2+2ab\cos\theta$....(I)

Magnitude of both vectors a and b

Let the angle which vector R make with vector a, be Φ

From vector algebra,

$\tan\phi=\dfrac{b\sin\theta}{a+b\cos\theta}$.....(II)

Given that,

$\phi=90^{\circ}$

$a+b\cos\theta=0$

$a=-b\cos\theta$.....(III)

Put the value of a in equation (I)

$R^2=b^2\cos^2\theta+b^2+2(-b\cos\theta)(b\cos\theta)$

$R^2=b^2+b^2\cos^2\theta-2b^2\cos^2\theta$

$R^2=b^2-b^2\cos^2\theta$

$R^2=b^2(1-\cos^2\theta)$

$R^2=b^2\sin^2\theta$

$(\dfrac{R}{b})^2=\sin^2\theta$

$\sin\theta=\dfrac{R}{b}$

Put the value of R

$\sin\theta=\dfrac{\dfrac{\sqrt{3}b}{2}}{b}$

$\sin\theta=\dfrac{\sqrt{3}}{2}$

$\theta=\sin^{-1}(\dfrac{\sqrt{3}}{2})$

$\theta=60^{\circ}$

Hence, The angle between them is 60°.

(A) is correct option.