Question

We have four vectors of equal magnitude as \[\overrightarrow{P},\overrightarrow{Q},\overrightarrow{R},\overrightarrow{S}\]. If \[\overrightarrow{P}+\overrightarrow{Q} -\overrightarrow{R}=0\] angle between \[\overrightarrow{P}\] and \[\overrightarrow{Q}\] is \[{{\theta }_{1}}\]. If \[\overrightarrow{P}+\overrightarrow{Q}-\overrightarrow{S}=0\] angle between \[\overrightarrow{P}\] and \[\overrightarrow{S}\] is \[{{\theta }_{2}}\]. The ratio of \[{{\theta }_{1}}\] to \[{{\theta }_{2}}\] is –
\[\begin{align}
  & \text{A) 1:2} \\
 & \text{B) 2:1} \\
 & \text{C) 1:1} \\
 & \text{D) 1:}\sqrt{3} \\
\end{align}\]

Answer 1

Hint: We have to relate the given vectors with each other to find the angle relation between the two given pairs of vectors. We can use the vector addition and magnitudes of the vectors to easily find the required angle between the two pairs of vectors.

Complete step by step solution:
We are given four-vectors \[\overrightarrow{P},\overrightarrow{Q},\overrightarrow{R},\overrightarrow{S}\] with equal magnitudes. We are given two combinations of these vectors such that their resultant becomes zero. Let us consider each situation one by one and find the angle involved to give the required solution.
We can find the square of each relation from which we will get the cosine relation between the two vectors. We can find the angle from this relation.
1. \[\overrightarrow{P}+\overrightarrow{Q}-\overrightarrow{R}=0\]: We can find the angle from the vector sum by squaring on both sides as –
\[\begin{align}
  & \overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R} \\
 & \Rightarrow {{\left| \overrightarrow{P} \right|}^{2}}+{{\left| \overrightarrow{Q} \right|}^{2}}+2\left| \overrightarrow{P} \right|\left| \overrightarrow{Q} \right|\cos {{\theta }_{1}}={{\left| \overrightarrow{R} \right|}^{2}} \\
\end{align}\]
We know that the magnitude of all the vectors are equal as per the conditions given in the problem. So, we can easily reduce the equation in terms of vector Q as –
\[\begin{align}
  & {{\left| \overrightarrow{P} \right|}^{2}}+{{\left| \overrightarrow{Q} \right|}^{2}}+2\left| \overrightarrow{P} \right|\left| \overrightarrow{Q} \right|\cos {{\theta }_{1}}={{\left| \overrightarrow{R} \right|}^{2}} \\
 & \Rightarrow 2{{\left| \overrightarrow{Q} \right|}^{2}}+2{{\left| \overrightarrow{Q} \right|}^{2}}\cos {{\theta }_{1}}={{\left| \overrightarrow{Q} \right|}^{2}} \\
 & \Rightarrow 2+2\cos {{\theta }_{1}}=1 \\
 & \Rightarrow \cos {{\theta }_{1}}=\dfrac{-1}{2} \\
 & \therefore {{\theta }_{1}}=\dfrac{2\pi }{3} \\
\end{align}\]
So, the angle between the vectors P and Q comes out to be \[\dfrac{2\pi }{3}\].
Now, let us consider the second situation, where the vector sum goes by –
\[\overrightarrow{P}+\overrightarrow{Q}-\overrightarrow{S}=0\]
We can continue the same method squaring the vector sum so as to ge the angle between vectors P and S this time. This can be given as –
\[\begin{align}
  & \overrightarrow{P}+\overrightarrow{Q}-\overrightarrow{S}=0 \\
 & \Rightarrow \overrightarrow{P}-\overrightarrow{S}=\overrightarrow{Q} \\
 & \text{Squaring on both sides,} \\
 & \Rightarrow {{\left| \overrightarrow{P} \right|}^{2}}+{{\left| \overrightarrow{S} \right|}^{2}}-2\left| \overrightarrow{P} \right|\left| \overrightarrow{S} \right|\cos {{\theta }_{2}}={{\left| \overrightarrow{Q} \right|}^{2}} \\
\end{align}\]
Now, we can consider all the vectors to be equal to vector Q and solve the angle as –
\[\begin{align}
  & {{\left| \overrightarrow{P} \right|}^{2}}+{{\left| \overrightarrow{S} \right|}^{2}}-2\left| \overrightarrow{P} \right|\left| \overrightarrow{S} \right|\cos {{\theta }_{2}}={{\left| \overrightarrow{Q} \right|}^{2}} \\
 & \Rightarrow 2{{\left| Q \right|}^{2}}-2{{\left| \overrightarrow{Q} \right|}^{2}}\cos {{\theta }_{2}}={{\left| \overrightarrow{Q} \right|}^{2}} \\
 & \Rightarrow 2-2\cos {{\theta }_{2}}=1 \\
 & \Rightarrow \cos {{\theta }_{2}}=\dfrac{-1}{-2} \\
 & \therefore {{\theta }_{2}}=\dfrac{\pi }{3} \\
\end{align}\]
The angle between the vectors P and Q turns out to be \[\dfrac{\pi }{3}\].
Now, we can find the ratio between the two angles as –
\[\begin{align}
  & \dfrac{{{\theta }_{1}}}{{{\theta }_{2}}}=\dfrac{\dfrac{2\pi }{3}}{\dfrac{\pi }{3}} \\
 & \therefore {{\theta }_{1}}:{{\theta }_{2}}=2:1 \\
\end{align}\]
The ratio between the angles \[{{\theta }_{1}}\text{ and }{{\theta }_{2}}\]is in the ratio 2:1.
The correct answer is option B.

Note: The vector sum of two or more vectors is completely dependent on the angle between them. The magnitude is just the factor, which determines how much the vector denotes. The direction between them can cancel off the huge vector quantities very easily.