Question

The vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ satisfy the equation $p\overrightarrow{x}+q\overrightarrow{y}=\overrightarrow{a}$ (where $p,q$ are scalar constants and $\overrightarrow{a}$ is a known vector). It is given that $\overrightarrow{x}.\overrightarrow{y}\ge \frac{|\overrightarrow{a}{|}^2}{4pq}$, then $\frac{|\overrightarrow{x}|}{|\overrightarrow{y}|}$ is equal to ($pq>0$)

• A. $1$
• B. $\frac{p^2}{q^2}$
• C. $\frac{p}{q}$
• D. $\frac{q}{p}$

Answer 1

The correct option is D $\frac{q}{p}$

$\overrightarrow{x}\cdot \overrightarrow{y}\ge \frac{|\overrightarrow{a}{|}^2}{4pq}$
$\Rightarrow |\overrightarrow{a}{|}^2-4pq\overrightarrow{x}\cdot \overrightarrow{y}\le 0\cdots \left(1\right)$

Also, $p\overrightarrow{x}+q\overrightarrow{y}=\overrightarrow{a}$
$|p\overrightarrow{x}+q\overrightarrow{y}|=|\overrightarrow{a}|$
Squaring both sides, we get
$p^2|\overrightarrow{x}{|}^2+2pq\overrightarrow{x}\cdot \overrightarrow{y}+q^2|\overrightarrow{y}{|}^2=|\overrightarrow{a}{|}^2$

$\Rightarrow p^2|\overrightarrow{x}{|}^2-2pq\overrightarrow{x}\cdot \overrightarrow{y}+q^2|\overrightarrow{y}{|}^2=|\overrightarrow{a}{|}^2-4pq\overrightarrow{x}\cdot \overrightarrow{y}$

$\Rightarrow |p\overrightarrow{x}-q\overrightarrow{y}{|}^2=|\overrightarrow{a}{|}^2-4pq\overrightarrow{x}\cdot \overrightarrow{y}$

Now,
$|p\overrightarrow{x}-q\overrightarrow{y}{|}^2\ge 0$
From Equation (1)
$|\overrightarrow{a}{|}^2-4pq\overrightarrow{x}\cdot \overrightarrow{y}\le 0|p\overrightarrow{x}-q\overrightarrow{y}{|}^2=0$

$\Rightarrow p|\overrightarrow{x}|=q|\overrightarrow{y}|$

$\Rightarrow \frac{|\overrightarrow{x}|}{|\overrightarrow{y}|}=\frac{q}{p}$