The vectors X- and Y- satisfy the equations 2X-Y-p- - X-2Y-q- where p-i-j- and q-i-j- lf - is the angle between X- and Y- then

The correct option is B cosθ= 45 #160;2 X⃗ + Y⃗ = #160;p⃗ = #160;i⃗ + #160;j⃗ #160; #160;....(1) #160;X⃗ + 2 #160;Y⃗ = #160;q⃗ = ...

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Byju's Answer

Standard XII

Mathematics

Pair of Lines Passing Though Origin

The vectors ...

Question

The vectors $\to X$ and $\to Y$ satisfy the equations $2 \to X + \to Y = \to p$ , $\to X + 2 \to Y = \to q$ where $\to p = \to i + \to j$ and $\to q = \to i - \to j$ . lf $θ$ 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 B $cos θ = - \frac{4}{5}$
$2 \to X + \to Y = \to p = \to i + \to j$ ....(1)
$\to X + 2 \to Y = \to q = \to i - \to j$ ....(2)
Multiplying equation $(2)$ by $2$ and subtracting equation $(1)$ from that, we get
$3 \to Y = \to i - 3 \to j; \Rightarrow \to Y = \frac{1}{3} (\to i - 3 \to j)$
And thus $\to X = \frac{1}{3} (\to i + 3 \to j)$
$∴ cos θ = \frac{\to X . \to Y}{| \to X | | \to Y |} = \frac{1 - 9}{10} = \frac{- 4}{5}$

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The vectors →X and →Y satisfy the equations 2→X+→Y=→p , →X+2→Y=→q where →p=→i+→j and →q=→i−→j. lf θ is the angle between →X and →Y, then

The correct option is B cosθ=−452→X+→Y=→p=→i+→j ....(1)→X+2→Y=→q=→i−→j ....(2)Multiplying equation (2) by 2 and subtracting equation (1) from that, we get3→Y=→i−3→j;⇒→Y=13(→i−3→j) And thus →X=13(→i+3→j)∴cosθ=→X.→Y|→X||→Y|=1−910=−45

The vectors $\to X$ and $\to Y$ satisfy the equations $2 \to X + \to Y = \to p$ , $\to X + 2 \to Y = \to q$ where $\to p = \to i + \to j$ and $\to q = \to i - \to j$ . lf $θ$ is the angle between $\to X$ and $\to Y$ , then