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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Mathematics
Chapter
Vector Algebra

JEE Advanced 2015 · Paper 1 · Mathematics · Vector Algebra

Let PQR be a triangle. Let → a = → QR , → b = → RP and → c = → PQ ....

Let δPQRbeatriangle.Let\delta PQR be a triangle. Let\overset{\to}{a} = \overset{\to}{QR}, \overset{\to}{b} = \overset{\to}{RP}and\overset{\to}{c} = \overset{\to}{PQ} .If∣a→∣=12,∣b→∣=43If \left|\overset{\to}{a}\right| = 12, \left|\overset{\to}{b}\right| = 4\sqrt{3} and b→⋅c→=24,\overset{\to}{b} ⋅ \overset{\to}{c} = 24, then which of the following is (are) true ?

Options

  1. A∣c∣→22−∣a→∣=12\dfrac{{\overset{\to}{\left|c\right|}}^{2}}{2} - \left|\overset{\to}{a}\right| = 12
  2. B∣c∣→22−∣a→∣=30\dfrac{{\overset{\to}{\left|c\right|}}^{2}}{2} - \left|\overset{\to}{a}\right| = 30
  3. C∣a→×b→+c→×a→∣=483\left|\overset{\to}{a} \times \overset{\to}{b} + \overset{\to}{c} \times \overset{\to}{a}\right| = 48\sqrt{3}
  4. D\overset{\to}{a} ⋅ \overset{\to}{b} = - 72

Correct answer

A. ∣c∣→22−∣a→∣=12\dfrac{{\overset{\to}{\left|c\right|}}^{2}}{2} - \left|\overset{\to}{a}\right| = 12

Step-by-step solution

\left|\overset{\to}{a}\right| = 12, \left|\overset{\to}{b}\right| = 4\sqrt{3}, \overset{\to}{b .}\overset{\to}{c} = 24<br><br >\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c} = 0(Sidesoftriangle)<br>(Sides of triangle)<br >\overset{\to}{b} + \overset{\to}{c} = - \overset{\to}{a}<br>(b→+c→).(b→+c→)=−a→.−a→<br > \left(\overset{\to}{b} + \overset{\to}{c}\right) . \left(\overset{\to}{b} + \overset{\to}{c}\right) = - \overset{\to}{a} . - \overset{\to}{a}
∣b→∣2+∣c→∣2+2b→.c→=∣a→∣2{\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} + 2\overset{\to}{b} . \overset{\to}{c} = {\left|\overset{\to}{a}\right|}^{2}
48+∣c→∣2+2×24=12248 + {\left|\overset{\to}{c}\right|}^{2} + 2 \times 24 = {12}^{2}
∣c→∣2=48{\left|\overset{\to}{c}\right|}^{2} = 48
∣c→∣22−∣a→∣=482−12=12\dfrac{{\left|\overset{\to}{c}\right|}^{2}}{2} - \left|\overset{\to}{a}\right| = \dfrac{48}{2} - 12 = 12
∣c→∣22+∣a→∣=482+12=36\dfrac{{\left|\overset{\to}{c}\right|}^{2}}{2} + \left|\overset{\to}{a}\right| = \dfrac{48}{2} + 12 = 36
\left|\overset{\to}{a} \times \overset{\to}{b} + \overset{\to}{c} \times \overset{\to}{a}\right| = \left|\overset{\to}{a} \times \overset{\to}{b} - \overset{\to}{a} \times \overset{\to}{c}\right|<br><br >= \left|\overset{\to}{a} \times \left(\overset{\to}{b} - \overset{\to}{c}\right)\right|<br><br >\overset{\to}{a} = - \overset{\to}{b} - \overset{\to}{c}<br><br >= \left|\left(- \overset{\to}{b} - \overset{\to}{c}\right) \times \left(\overset{\to}{b} - \overset{\to}{c}\right)\right| = \left|\left(\overset{\to}{b} - \overset{\to}{c}\right) \times \left(\overset{\to}{b} + \overset{\to}{c}\right)\right|<br><br >= 2\left|\overset{\to}{b} \times \overset{\to}{c}\right|<br><br >= 2\sqrt{{\left|\overset{\to}{b}\right|}^{2}{\left|\overset{\to}{c}\right|}^{2} - {\left(\overset{\to}{b} . \overset{\to}{c}\right)}^{2}} = 2\sqrt{48 \times 48 - {\left(24\right)}^{2}}<br><br >= 2 \times 24\sqrt{4 - 1} = 48\sqrt{3}<br><br >\overset{\to}{a} + \overset{\to}{b} = - \overset{\to}{c}<br>(a→+b→).(a→+b→)=−c→.−c→<br > \left(\overset{\to}{a} + \overset{\to}{b}\right) . \left(\overset{\to}{a} + \overset{\to}{b}\right) = - \overset{\to}{c} . - \overset{\to}{c}
∣a→∣2+∣b→∣2+2a→.b→=∣c→∣2{\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + 2\overset{\to}{a} . \overset{\to}{b} = {\left|\overset{\to}{c}\right|}^{2}
a→.b→=−∣a→∣22=−1222=−72\overset{\to}{a} . \overset{\to}{b} = \dfrac{- {\left|\overset{\to}{a}\right|}^{2}}{2} = - \dfrac{{12}^{2}}{2} = - 72
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