The angle between P→ and Q→ is cos−1(−3b2a),if |P→|=|a→+2b→|. Find Q,→ if |P→|=|Q→|:

cosα=−3b2a=−3b22abWhich gives 3b2+2abcosα=0Given, |P→|2=a2+4b2+4abcosαAnd |P→|2=|Q→|2Equation (1) can be achieved if |P→|2=a2+4b2+4abcosα=a2+b2+2abcosαSo, |Q→|2=a2+b2+2abcosαQ→=(a→+b→)

The angle between P→ and Q→ is cos−1(−3b2a),if |P→|=|a→+2b→|. Find Q,→ if |P→|=|Q→|:

cosα=−3b2a=−3b22abWhich gives 3b2+2abcosα=0Given, |P→|2=a2+4b2+4abcosαAnd |P→|2=|Q→|2Equation (1) can be achieved if |P→|2=a2+4b2+4abcosα=a2+b2+2abcosαSo, |Q→|2=a2+b2+2abcosαQ→=(a→+b→)

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The angle between $\vec{P}$ and $\vec{Q}$ is $\cos^{- 1} (- \frac{3 b}{2 a})$ ,if $| \vec{P} | = | \vec{a} + \vec{2 b} | .$ Find $\vec{Q,}$ if $| \vec{P} | = | \vec{Q} | :$

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$(\vec{a} - \vec{2 b})$

$(2 \vec{a} + \vec{b})$

$(\vec{a} + \vec{b})$

$(2 \vec{a} - 2 \vec{b})$

answer is C.

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Detailed Solution

$\cos α = \frac{- 3 b}{2 a} = - \frac{3 b^{2}}{2 a b}$
Which gives
$3 b^{2} + 2 a b \cos α = 0$
Given, ${| \vec{P} |}^{2} = a^{2} + 4 b^{2} + 4 a b \cos α$
And ${| \vec{P} |}^{2} = {| \vec{Q} |}^{2}$
Equation (1) can be achieved if
${| \vec{P} |}^{2} = a^{2} + 4 b^{2} + 4 a b \cos α$
$= a^{2} + b^{2} + 2 a b \cos α$
So, ${| \vec{Q} |}^{2} = a^{2} + b^{2} + 2 a b \cos α$
$\vec{Q} = (\vec{a} + \vec{b})$