Question

For two vectors $\overrightarrow{A}$ and $\overrightarrow{B},$ sum $\overrightarrow{A}+\overrightarrow{B}$ is perpendicular to the difference $\overrightarrow{A}-\overrightarrow{B}$. Then the ratio of their magnitudes may be

• A. 1
• B. 2
• C. 3
• D. none of these

Answer 1

The correct option is A 1

$(\overrightarrow{A}+\overrightarrow{B})$ is perpendicular to $(\overrightarrow{A}-\overrightarrow{B}).$
Thus $(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}-\overrightarrow{B})=0$
$\overrightarrow{A}\cdot \overrightarrow{A}-\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{B}\cdot \overrightarrow{A}-\overrightarrow{B}\cdot \overrightarrow{B}=0$
By commutative property of dot product
$\overrightarrow{A}.\overrightarrow{B}=\overrightarrow{B}.\overrightarrow{A}$
$\therefore A^2-B^2=0$ or $A=\pm B$

Thus the ratio of magnitudes may be
$\frac{A}{B}=1$ (since magnitude cannot be negative)

Alternative: According to parallelogram law of vector addition, $\overrightarrow{A}+\overrightarrow{B}$ and $\overrightarrow{A}-\overrightarrow{B}$ are diagonals.
Because diagonals are perpendicular according to question, it is a rhombus. Hence the ratio of magnitudes of the two vectors is 1.