Question

the resultant of two vectors p bar and q bar is perpendicular to p bar, and it's magnitude is half that q bar. what is the angle between p bar and q bar?​

Answer 1

Solution:-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\put(0,0){\vector(3,0){3}}\put(0,0){\vector(-3,4){2}}\put(0,0){\vector(0,3){3}}\put(2,-0.5){{\footnotesize \vec{p} }}\put(-2,2.1){{\footnotesize \vec{q} }}\put(0.5,2){{\footnotesize \vec{r} }}\put(0,0){\vector(3,4){2}}\put(1,1.){{\footnotesize \vec{r+p} }}\end{picture}$

Let the magnitude of $\vec{p}$ be p and magnitude of $\vec{q}$ be q

According to the question,

The magnitude of $\vec{r}$ is half that of the magnitude of $\vec{q}$

Therefore,

Magnitude of $\vec{r}$ = q/2

We know that when two vectors are perpendicular then the magnitude of their difference as well as the sum is equal.

Here the difference of the vectors is equal to $\vec{q}$

Therefore,

q = q/2 + p

p = q/2

We know that the magnitude of the resultant vector can be found out using the formula,

$\rm R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$, $\theta$ is the angle between the two vectors

Therefore the magnitude of $\vec{r}$ is,

$\rm r = \biggg{\sqrt{\frac{q^2}{4}+\frac{q^2}{4}+2\times\frac{q^2}{4}\times \cos\theta }}$

$\rm r =\bigg{ \sqrt{\frac{q^2}{2}+\frac{q^2\cos\theta}{2} }}$

But r = q/2

Therefore,

$\dfrac{q^2}{4} = \dfrac{q^2}{2} + \dfrac{q^2\cos\theta}{2}$

$\dfrac{q^2 - 2q^2}{2} = \dfrac{q^2\cos\theta}{2} \implies \dfrac{-q^2}{2} = \dfrac{q^2\cos\theta}{2} \implies \cos\theta = - \dfrac{1}{2}$$\theta = {\cos}^{-1}(1/2) \implies \theta = 120^{\circ}$