Verified Solution • 2025–26 Session moderate

QJEEAdv13Mace4b: JEE Advanced 2013 Physics - Class 11 JEEAdv13Mace4b Solution: match List with List and select the correct answer using the…

match List II with List IIII and select the correct answer using the code given below the lists:

    \,\,\,\,     \,\,\,\,     \,\,\,\, List II
(P.)     \,\,\,\, Volume of parallelopiped determined by vectors a→,b→\overrightarrow a ,\overrightarrow b and c→\overrightarrow c is 2.2. Then the volume of the parallelepiped determined by vectors 2(a→×b→),3(b→×c→)2\left( {\overrightarrow a \times \overrightarrow b } \right),3\left( {\overrightarrow b \times \overrightarrow c } \right) and (c→×a→)\left( {\overrightarrow c \times \overrightarrow a } \right) is
(Q.)     \,\,\,\, Volume of parallelopiped determined by vectors a→,b→\overrightarrow a ,\overrightarrow b and c→\overrightarrow c is 5.5. Then the volume of the parallelepiped determined by vectors 3(a→+b→),(b→+c→)3\left( {\overrightarrow a + \overrightarrow b } \right),\left( {\overrightarrow b + \overrightarrow c } \right) and 2(c→+a→)2\left( {\overrightarrow c + \overrightarrow a } \right) is
(R.)     \,\,\,\, Area of a triangle with adjacent sides determined by vectors a→{\overrightarrow a } and b→{\overrightarrow b } is 20.20. Then the area of the triangle with adjacent sides determined by vectors (2a→+3b→)\left( {2\overrightarrow a + 3\overrightarrow b } \right) and (a→−b→)\left( {\overrightarrow a - \overrightarrow b } \right) is
(S.)     \,\,\,\, Area of a parallelogram with adjacent sides determined by vectors a→{\overrightarrow a } and b→{\overrightarrow b } is 30.30. Then the area of the parallelogram with adjacent sides determined by vectors (a→+b→)\left( {\overrightarrow a + \overrightarrow b } \right) and a→{\overrightarrow a } is

    \,\,\,\,     \,\,\,\,     \,\,\,\, List IIII
(1.)     \,\,\,\, 100100
(2.)     \,\,\,\, 3030
(3.)     \,\,\,\, 2424
(4.)     \,\,\,\, 6060

Multiple Choice Options

  • A P=4,Q=2,R=3,S=1P = 4,Q = 2,R = 3,S = 1
  • B P=2,Q=3,R=1,S=4P = 2,Q = 3,R = 1,S = 4
  • C P=3,Q=4,R=1,S=2P = 3,Q = 4,R = 1,S = 2
  • D P=1,Q=4,R=3,S=2P = 1,Q = 4,R = 3,S = 2

Step-by-Step Verified Solution

(P) Given, the volume of parallelopiped formed by a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} is 2

⇒[a⃗b⃗c⃗]=2\Rightarrow[\vec{a} \vec{b} \vec{c}]=2

Let V be the volume of parallelopiped formed by 2(a⃗×b⃗),3(b⃗×c⃗)2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c}) and (c⃗×a⃗)(\vec{c} \times \vec{a})

amp;⇒V=[2(a⃗×b⃗)amp;3(b⃗×c⃗)amp;(c⃗×a⃗)]amp;⇒V=2×3[(a⃗×b⃗)(b⃗×c⃗)(c⃗×a⃗)]amp;⇒V=6[a⃗b⃗c⃗]2amp;⇒V=6×22amp;⇒V=24\begin{aligned}& \Rightarrow \mathrm{V}=\left[\begin{array}{lll}2(\vec{a} \times \vec{b}) & 3(\vec{b} \times \vec{c}) & (\vec{c} \times \vec{a})\end{array}\right] \\& \Rightarrow \mathrm{V}=2 \times 3[(\vec{a} \times \vec{b})(\vec{b} \times \vec{c})(\vec{c} \times \vec{a})] \\& \Rightarrow \mathrm{V}=6[\vec{a} \vec{b} \vec{c}]^2 \\& \Rightarrow \mathrm{V}=6 \times 2^2 \\& \Rightarrow \mathrm{V}=24 \\\end{aligned}

Hence, P match with 3.

(Q) Given, the volume of parallelopiped formed by a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} is 5.

⇒[a⃗b⃗c⃗]=5... (i)\Rightarrow[\vec{a} \vec{b} \vec{c}]=5 \quad \text{... (i)}

Let V be the volume of parallelopiped formed by 3(a⃗+b⃗),(b⃗+c⃗)3(\vec{a}+\vec{b}),(\vec{b}+\vec{c}) and 2(c⃗+a⃗)2(\vec{c}+\vec{a}) .

⇒V=[3(a⃗+b⃗)amp;(b⃗+c⃗)amp;2(c⃗+a⃗)]\Rightarrow \mathrm{V}=\left[\begin{array}{lll}3(\vec{a}+\vec{b}) & (\vec{b}+\vec{c}) & 2(\vec{c}+\vec{a})\end{array}\right]

amp;⇒V=3.2[(a⃗+b⃗)(b⃗+c⃗)(c⃗+a⃗)]amp;⇒V=6×2[a⃗amp;b⃗amp;c⃗]amp;⇒V=12×5amp;⇒V=60\begin{aligned}& \Rightarrow \mathrm{V}=3.2[(\vec{a}+\vec{b})(\vec{b}+\vec{c})(\vec{c}+\vec{a})] \\& \Rightarrow \mathrm{V}=6 \times 2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \\& \Rightarrow \mathrm{V}=12 \times 5 \\& \Rightarrow \mathrm{V}=60\end{aligned}

Hence, Q match with 4.

(R) Given, the area of a triangle with adjacent sides a⃗\vec{a} and b⃗\vec{b} is 20 .

⇒amp;12∣a⃗×b⃗∣=20⇒amp;∣a⃗×b⃗∣=40.... (i)\begin{array}{ll}\Rightarrow & \frac{1}{2}|\vec{a} \times \vec{b}|=20 \\\Rightarrow & |\vec{a} \times \vec{b}|=40 \quad \text{.... (i)}\end{array}

Let Δ\Delta be the area of a triangle with adjacent sides (2a⃗+3b⃗)(2 \vec{a}+3 \vec{b}) and (a⃗−b⃗)(\vec{a}-\vec{b}) .

amp;⇒Δ=12∣(2a⃗+3b⃗)×(a⃗−b⃗)∣amp;⇒Δ=12∣2a⃗×a⃗−2a⃗×b⃗+3b⃗×a⃗−3b⃗×b⃗∣amp;⇒Δ=12∣0−2a⃗×b⃗−3a⃗×b⃗−0∣amp;⇒Δ=52∣a⃗×b⃗∣amp;⇒Δ=52×40amp;⇒Δ=100\begin{aligned}& \Rightarrow \quad \Delta=\frac{1}{2}|(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})| \\& \Rightarrow \Delta=\frac{1}{2}|2 \vec{a} \times \vec{a}-2 \vec{a} \times \vec{b}+3 \vec{b} \times \vec{a}-3 \vec{b} \times \vec{b}| \\& \Rightarrow \Delta=\frac{1}{2}|0-2 \vec{a} \times \vec{b}-3 \vec{a} \times \vec{b}-0| \\& \Rightarrow \Delta=\frac{5}{2}|\vec{a} \times \vec{b}| \\& \Rightarrow \Delta=\frac{5}{2} \times 40 \\& \Rightarrow \quad \Delta=100\end{aligned}

Hence, R match with 1.

(S) Given, the area of parallelogram with adjacent sides a⃗\vec{a} and b⃗\vec{b} is 30.

⇒∣a⃗×b⃗∣=30... (i)\Rightarrow|\vec{a} \times \vec{b}|=30 \quad \text{... (i)}

Let, Δ′\Delta^{\prime} be the area of parallelogram with adjacent sides (a⃗+b⃗)(\vec{a}+\vec{b}) and a⃗\vec{a} .

⇒amp;Δ′=∣(a⃗+b⃗)×a⃗∣⇒amp;Δ′=∣a⃗×a⃗+b⃗×a⃗∣⇒amp;Δ′=∣0−a⃗×b⃗∣⇒amp;Δ′=∣a⃗×b⃗∣⇒amp;Δ′=30\begin{array}{ll}\Rightarrow & \Delta^{\prime}=|(\vec{a}+\vec{b}) \times \vec{a}| \\\Rightarrow & \Delta^{\prime}=|\vec{a} \times \vec{a}+\vec{b} \times \vec{a}| \\\Rightarrow & \Delta^{\prime}=|0-\vec{a} \times \vec{b}| \\\Rightarrow & \Delta^{\prime}=|\vec{a} \times \vec{b}| \\\Rightarrow & \Delta^{\prime}=30\end{array}

Hence, S match with 2.

Hints :

(i) [a⃗×b⃗b⃗×c⃗c⃗×a⃗]=[a⃗b⃗c⃗]2[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=[\vec{a} \vec{b} \vec{c}]^2

(ii) [a⃗+b⃗amp;b⃗+c⃗amp;c⃗+a⃗]=2[a⃗b⃗c⃗]\left[\begin{array}{lll}\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}\end{array}\right]=2\left[\begin{array}{ll}\vec{a} \vec{b} \vec{c}\end{array}\right]

(iii) A⃗⋅(B⃗×C⃗)=[A⃗B⃗C⃗]\vec{A} \cdot(\vec{B} \times \vec{C})=[\vec{A} \vec{B} \vec{C}]

(iv) A⃗×B⃗=−B⃗×A⃗\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}

(v) A⃗×A⃗=B⃗×B⃗=0\vec{A} \times \vec{A}=\vec{B} \times \vec{B}=0

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