The vectors x and y satisfy the equation px-qy-a where p- q are scalar constants and a is a known vector- It is given that x-y-a-24pq- then -x-y- is equal to pq0

X·y≥|a|^24pq ⇒ |a|^2 4pq x·y≤ 0 ⋯(1) Also, px+qy=a |px+qy|=|a| Squaring both sides, we get p^2|x|^2+2pqx·y+q^2|y|^2=|a|^2 ⇒ p^2|x|^2 2pqx·y+q^2|y|^2=|a|^2 4 ...

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Byju's Answer

Standard XIII

Mathematics

Applications of Dot Product

The vectors x...

Question

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

1

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\frac{p^{2}}{q^{2}}

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\frac{p}{q}

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\frac{q}{p}

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Solution

The correct option is D $\frac{q}{p}$
$\to x \cdot \to y \geq \frac{| \to a |^{2}}{4 p q}$
$\Rightarrow | \to a |^{2} - 4 p q \to x \cdot \to y \leq 0 \dots (1)$

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

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

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

Now,
$| p \to x - q \to y |^{2} \geq 0$
From Equation (1)
$| \to a |^{2} - 4 p q \to x \cdot \to y \leq 0 | p \to x - q \to y |^{2} = 0$

$\Rightarrow p | \to x | = q | \to y |$

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

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The vectors →x and →y satisfy the equation p→x+q→y=→a (where p, q are scalar constants and →a is a known vector). It is given that →x.→y≥|→a|24pq, then |→x||→y| is equal to (pq>0)

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