Question

The sum and the difference of the two perpendicular vectors of equal lengths are:
(A) Of equal lengths and have an acute angle between them
(B) Of equal length and have an obtuse angle between them
(C) Also perpendicular to each other and are of different lengths
(D) Also perpendicular to each other and are of equal lengths

Answer 1

Hint: The resultant of sum and difference of two vectors is given by:
 $\Rightarrow \overrightarrow{A}\pm \overrightarrow{B}=\sqrt{{{A}^{2}}+{{B}^{2}}\pm 2AB\cos \theta } $
And the angle between two resultant vectors is given by:
 $\Rightarrow \cos \theta =\dfrac{\left( \overrightarrow{A}+\overrightarrow{B} \right)\left( \overrightarrow{A}-\overrightarrow{B} \right)}{{{\left| \left( \overrightarrow{A}+\overrightarrow{B} \right) \right|}^{2}}{{\left| \left( \overrightarrow{A}-\overrightarrow{B} \right) \right|}^{2}}} $

Complete step by step solution
Considering two vectors $ \overrightarrow{A}\text{ and }\overrightarrow{B} $ which are perpendicular to each other and are of equal length.
Their sum is given by:
 $ \begin{align}
  &\Rightarrow Sum=\overrightarrow{A}+\overrightarrow{B} \\
 &\Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta } \\
 &\Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos 90{}^\circ } \\
 &\Rightarrow Sum=\sqrt{{{A}^{2}}+{{B}^{2}}}\text{ }............................\text{ (1)} \\
\end{align} $
And the difference is given by:
 $ \begin{align}
 &\Rightarrow Difference=\overrightarrow{A}-\overrightarrow{B} \\
 &\Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta } \\
 &\Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos 90{}^\circ } \\
 &\Rightarrow Difference=\sqrt{{{A}^{2}}+{{B}^{2}}}\ \text{ }...............................\text{ (2)} \\
\end{align} $
Comparing equations (1) and (2),
We can clearly see that the resultants have equal lengths.
Now calculating the angle between them:
 $ \begin{align}
 &\Rightarrow \cos \theta =\dfrac{\left( \overrightarrow{A}+\overrightarrow{B} \right)\left( \overrightarrow{A}-\overrightarrow{B} \right)}{{{\left| \left( \overrightarrow{A}+\overrightarrow{B} \right) \right|}^{2}}{{\left| \left( \overrightarrow{A}-\overrightarrow{B} \right) \right|}^{2}}} \\
 &\Rightarrow \cos \theta =\dfrac{\left( \overrightarrow{A}+\overrightarrow{B} \right)\left( \overrightarrow{A}-\overrightarrow{B} \right)}{\left| {{A}^{2}}+{{B}^{2}} \right|\left| {{A}^{2}}-{{B}^{2}} \right|} \\
 &\Rightarrow \cos \theta =\dfrac{{{A}^{2}}-{{B}^{2}}}{{{A}^{2}}+{{B}^{2}}} \\
 &\Rightarrow \cos \theta =0 \\
 &\Rightarrow \theta =90{}^\circ \\
\end{align} $
Therefore, the two resultant vectors are perpendicular to each other.
Hence, option (D) is the correct answer.

Note
We can also find the angle between the two resultant vectors in a simplified way:
 $\Rightarrow As\ \text{ }\overrightarrow{A}=\overrightarrow{B} $
Squaring both sides we get;
 $\Rightarrow {{\left( \overrightarrow{A} \right)}^{2}}={{\left( \overrightarrow{B} \right)}^{2}} $
Or
 $\Rightarrow {{\left( \overrightarrow{A} \right)}^{2}}-{{\left( \overrightarrow{B} \right)}^{2}}=0 $
We know, $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $
So using this identity we get;
 $\Rightarrow \left( \overrightarrow{A}+\overrightarrow{B} \right)\cdot \left( \overrightarrow{A}-\overrightarrow{B} \right)=0 $
As dot product of a vector is given by $ a\cdot b=ab\cos \theta $
Here
  $ \begin{align}
  &\Rightarrow a=\overrightarrow{A}+\overrightarrow{B}\text{ }(Sum) \\
 & \text{And} \\
 &\Rightarrow b=\overrightarrow{A}-\overrightarrow{B}\text{ }(Difference) \\
\end{align} $
As we can see that the dot product is zero where a and b cannot be zero so
 $ \begin{align}
 &\Rightarrow \cos \theta =0 \\
 & \therefore ,\ \theta =90{}^\circ \\
\end{align} $ .