The vectors x and y satisfy the equations 2x-y-p and x-2y-q- where p- and q- - If - is the angle between x and y then

The correct option is D cosθ = 45 Given equations 2x #160;+ y = P #160; #160; #160; #160; .....(1) x #160;+2y = q #160;.. ...

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Standard X

Mathematics

Section Formula

The vectors ...

Question

The vectors $\to x$ and $\to y$ satisfy the equations $2 \to x + \to y = \to p$ and $\to x + 2 \to y = \to q$ , where $\to p =^i +^j$ and $\to q =^i -^j$ . If $θ$ is the angle between $\to x$ and $\to y$ then

cos θ = \frac{4}{5}

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sin θ = \frac{1}{\sqrt{2}}

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cos θ = - \frac{4}{5}

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cos θ = - \frac{3}{5}

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Solution

The correct option is D $cos θ = - \frac{4}{5}$
Given equations
$2 \to x + \to y = \to P$ .....(1)
$\to x + 2 \to y = \to q$ ....(2)
Solving (1) and (2), we get
$\frac{2 \to p - \to q}{3} = \to x$
$\frac{2 \to q - \to p}{3} = \to y$
Also given $\to p =^i +^j$ and $\to q =^i -^j$
$\Rightarrow \to x = \frac{^i + 3^j}{3}$
$\to y = \frac{^i - 3^j}{3}$
$cos θ = \frac{\to x \cdot \to y}{| \to x | | \to y |}$
$cos θ = \frac{(^i + 3^j) \cdot (^i - 3^j)}{\sqrt{10} \sqrt{10}}$
$= \frac{- 8}{10}$
$\Rightarrow cos θ = \frac{- 4}{5}$

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The vectors →x and →y satisfy the equations 2→x+→y=→p and →x+2→y=→q, where →p=^i+^j and →q=^i−^j. If θ is the angle between →x and →y then

The vectors $\to x$ and $\to y$ satisfy the equations $2 \to x + \to y = \to p$ and $\to x + 2 \to y = \to q$ , where $\to p =^i +^j$ and $\to q =^i -^j$ . If $θ$ is the angle between $\to x$ and $\to y$ then