Question

The magnitude of the resultant of the two vectors of magnitude 4 and 3 is 1. The angle between the vectors is:
A. 0
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{2}$
D. $\pi$

Answer 1

In the question, we are given two vectors having different magnitudes. The magnitude of vectors is calculated from the cosine and sine coefficients of that vector. The resultant of these vectors is 1. The vectors will have some angle in them which can be calculated from the formula of resultant given by:
$\operatorname{Re} = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \\\ \theta \,is\,the\,angle \\\$

Complete step-by-step solution:
Given, vectors have magnitude 4 and 3
Let $\left| A \right| = 4,\left| B \right| = 3$
Resultant of the two vectors is 1
From the formula
$\operatorname{Re} = \sqrt {{A^2} + {B^2} + 2AB\cos \theta }$
We have A, B and Resultant we need to find out the angle between vectors
Substituting the given values in the equation we get.
$\operatorname{Re} = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \\\ \Rightarrow 1 = \sqrt {{4^2} + {3^2} + 2 \times 4 \times 3\cos \theta } \\\$
Squaring on both sides we get
$\Rightarrow 1 = {4^2} + {3^2} + 2 \times 4 \times 3\cos \theta \\\ \Rightarrow 1 = 16 + 9 + 24\cos \theta \\\ \\\$
Further solving the equation and simplifying we get,
$\Rightarrow 24\cos \theta = - 24 \\\ \Rightarrow \cos \theta = ( - 1) \\\ \Rightarrow \theta = \pi \\\$
The angle between the vectors of magnitude 4 and 3 is $\pi$
Hence, option (D) is correct.

Note: The angle between vectors can be anything between 0 to $2\pi$ . We need to understand the values of cosines in different quadrants of the x-y plane. Cosine is positive in ${1^{st}},{3^{rd}}$ quadrant so the values in ${2^{nd}},{4^{th}}$ quadrants will be negative.