Question

If for two vector $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$, sum $\left(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\right)$ is perpendicular to the difference $\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right)$. The ratio of their magnitude is

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is A

$1$

Explanation for the correct answer:

The dot product of two vectors when they are perpendicular to each other is $0$

We know that the sum $\left(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\right)$ is perpendicular to the difference $\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right)$,

$\begin{array}{rcl}\therefore \left(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\right).\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right)& =& 0\\ & \Rightarrow & {\mathrm{A}}^2+\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}-{\mathrm{B}}^2=0\end{array}$

The dot product of two vectors is commutative which means when there are two vectors $\hat{a}\text{and}\hat{b}$ then $\hat{a}·\hat{b}=\hat{b}·\hat{a}$. Apply this to the above equation,

$\begin{array}{rcl}\therefore {\mathrm{A}}^2-{\mathrm{B}}^2& =& 0\\ & \Rightarrow & {\mathrm{A}}^2={\mathrm{B}}^2\end{array}$

After taking the square on both sides, the ratio of magnitude is

$\frac{\mathrm{A}}{\mathrm{B}}=1$.

Hence, the correct option is (A).