Verified Solution • 2025–26 Session moderate

QJEEM24Maa790: JEE Main 2024 Physics - Class 11 JEEM24Maa790 Solution: Let . Let be the vector such that and . Then is equal to :

Let a→=i^+2j^+k^,\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k},
b→=3(i^−j^+k^)\overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k}) .
Let c→\overrightarrow{\mathrm{c}} be the vector such that a→×c→=b→\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} and a⃗⋅c⃗=3\vec{a} \cdot \vec{c}=3 .
Then a⃗⋅((c⃗×b⃗)−b⃗−c⃗)\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c}) is equal to :

Multiple Choice Options

  • A 32
  • B 36
  • C 24
  • D 20

Step-by-Step Verified Solution

amp;a⃗⋅[(c⃗×b⃗)−b⃗−c⃗]amp;a⃗⋅(c⃗×b⃗)−a⃗⋅b⃗−a⃗⋅c⃗..... (i)\begin{aligned}& \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\& \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)}\end{aligned}

amp; given a⃗×c⃗=b⃗amp;⇒(a⃗×c⃗)⋅b⃗=b⃗⋅b⃗=∣b⃗∣2=27amp;⇒a⃗⋅(c⃗×b⃗)=[a⃗c⃗b⃗]=(a⃗×c⃗)⋅b⃗=27.... (ii)\begin{aligned}& \text { given } \vec{a} \times \vec{c}=\vec{b} \\& \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\& \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)}\end{aligned}

amp; Now a⃗⋅b⃗=3−6+3=0.... (iii)amp;a⃗⋅c⃗=3.... (iv) (given)\begin{aligned}& \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\& \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)}\end{aligned}

 By (i), (ii), (iii) & (iv) 27−0−3=24

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