Question

Angle between two vectors of magnitudes $12$ and $18$ units, when their resultant is $24$ units is

• A. $82^{\circ }31^{{}^{\prime }}$
• B. $63^{\circ }51^{{}^{\prime }}$
• C. $89^{\circ }16^{{}^{\prime }}$
• D. $75^{\circ }52^{{}^{\prime }}$.

Answer 1

Answer: (D)

Given: Magnitude of first vector $(\vec{A})=12$;
Magnitude of second vector $(\vec{B})=18$
and resultant of the given vectors $(\vec{R})=24$.
We know that resultant vector $|\vec{R}|=24=\sqrt{A^2+B^2+2AB\cos \theta }$
$=\sqrt{(12{)}^2+(18{)}^2+2\times 12\times 18\cos }$
or $(24{)}^2=144+324+432\cos \theta$
$\Rightarrow 432\cos \theta =108$
$\Rightarrow \cos \theta =\frac{108}{432}=0\times 25$
or $\theta ={\cos }^{-1}0\times 25=75^{\circ }52^{\prime }.$