Verified Solution • 2025–26 Session moderate

QJEEAdv15Ma25ef: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ma25ef Solution: Let be a triangle. Let and If then which of the following is…

Let ΔPQR\Delta PQR be a triangle. Let a⃗=QR→,b⃗=RP→\vec a = \overrightarrow {QR} ,\vec b = \overrightarrow {RP} and c→=PQ→.\overrightarrow c = \overrightarrow {PQ} . If ∣a→∣=12,  ∣b→∣=43,   b→.c→=24,\left| {\overrightarrow a } \right| = 12,\,\,\left| {\overrightarrow b } \right| = 4\sqrt 3 ,\,\,\,\overrightarrow b .\overrightarrow c = 24, then which of the following is (are) true?

Multiple Choice Options

  • A ∣c→∣22−∣a→∣=12{{{{\left| {\overrightarrow c } \right|}^2}} \over 2} - \left| {\overrightarrow a } \right| = 12
  • B ∣c→∣22+∣a→∣=30{{{{\left| {\overrightarrow c } \right|}^2}} \over 2} + \left| {\overrightarrow a } \right| = 30
  • C ∣a→×b→+c→×a→∣=483\left| {\overrightarrow a \times \overrightarrow b + \overrightarrow c \times \overrightarrow a } \right| = 48\sqrt 3
  • D a→.b→=−72\overrightarrow a .\overrightarrow b = - 72

Step-by-Step Verified Solution

For a triangle, we have a⃗+b⃗+c⃗=0→\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}

JEE Advanced 2015 Paper 1 Offline Mathematics - Vector Algebra Question 49 English Explanation
amp;⇒a⃗=−(b⃗+c⃗)amp;⇒∣a⃗∣2=∣b⃗∣2+∣c⃗∣2+2b⃗⋅c⃗amp;⇒∣a⃗∣2=48+∣c⃗∣2+48amp;⇒∣c⃗∣2=∣a⃗∣2−96=144−96amp;⇒∣c⃗∣2=48amp;⇒∣c⃗∣=48=43\begin{aligned}& \Rightarrow \vec{a}=-(\vec{b}+\vec{c}) \\\\& \Rightarrow|\vec{a}|^2=|\vec{b}|^2+|\vec{c}|^2+2 \vec{b} \cdot \vec{c} \\\\& \Rightarrow|\vec{a}|^2=48+|\vec{c}|^2+48 \\\\& \Rightarrow|\vec{c}|^2=|\vec{a}|^2-96=144-96 \\\\& \Rightarrow|\vec{c}|^2=48 \\\\& \Rightarrow|\vec{c}|=\sqrt{48}=4 \sqrt{3}\end{aligned}

Therefore, ∣c⃗∣22−∣a⃗∣=24−12=12\frac{|\vec{c}|^2}{2}-|\vec{a}|=24-12=12

Option (A) is correct.

∣c⃗∣22+∣a⃗∣=24+12=36\frac{|\vec{c}|^2}{2}+|\vec{a}|=24+12=36

Also, a⃗+b⃗=−c⃗⇒a⃗⋅b⃗=−72\quad \vec{a}+\vec{b}=-\vec{c} \Rightarrow \vec{a} \cdot \vec{b}=-72

From Eq. (1), we get

amp;a⃗×a⃗=−(a⃗×b⃗+a⃗×c⃗)⇒a⃗×b⃗=c⃗×a⃗amp;∣a⃗×b⃗+a⃗×c⃗∣=∣2(a⃗×b⃗)∣=2∣a⃗∣∣b⃗∣sin⁡θamp;=9631−(−72483)2amp;=9631−(312)2=483\begin{aligned}& \vec{a} \times \vec{a}=-(\vec{a} \times \vec{b}+\vec{a} \times \vec{c}) \Rightarrow \vec{a} \times \vec{b}=\vec{c} \times \vec{a} \\\\& |\vec{a} \times \vec{b}+\vec{a} \times \vec{c}|=|2(\vec{a} \times \vec{b})|=2|\vec{a}||\vec{b}| \sin \theta \\\\& =96 \sqrt{3} \sqrt{1-\left(\frac{-72}{48 \sqrt{3}}\right)^2} \\\\& =96 \sqrt{3} \sqrt{1-\left(\frac{\sqrt{3}}{12}\right)^2}=48 \sqrt{3}\end{aligned}

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