Verified Solution • 2025–26 Session moderate

QJEEM23Maa412: JEE Main 2023 Physics - Class 11 JEEM23Maa412 Solution: Let and . Let be a vector which is perpendicular to both and ,…

Let a⃗=2i^+7j^−k^,b⃗=3i^+5k^\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k} and c⃗=i^−j^+2k^\vec{c}=\hat{i}-\hat{j}+2 \hat{k} . Let d⃗\vec{d} be a vector which is perpendicular to both a⃗\vec{a} and b⃗\vec{b} , and c⃗⋅d⃗=12\vec{c} \cdot \vec{d}=12 . Then (−i^+j^−k^)⋅(c⃗×d⃗)(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) is equal to :

Multiple Choice Options

  • A 24
  • B 42
  • C 44
  • D 48

Step-by-Step Verified Solution

If d⃗\vec{d} is ⊥\perp to both a⃗\vec{a} and b⃗\vec{b} then

d⃗=λ(a⃗×b⃗)=λ∣i^amp;j^amp;k^2amp;7amp;−13amp;0amp;5∣=(35i^−13j^−21k^)λ\vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\2 & 7 & -1 \\3 & 0 & 5\end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda

amp; but c⃗⋅d⃗=12⇒λ(35×1+13×1−21×2)=12amp;⇒λ(6)=12⇒λ=2amp;λ⃗=2(35i^−13j^−21k^)\begin{aligned}& \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\& \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\& \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k})\end{aligned}

amp; Now, (−i^+j^−k^)⋅(c⃗×d⃗)amp;=∣−1amp;1amp;−11amp;−1amp;270amp;−26amp;−42∣amp;=−94+182−44=44\begin{aligned}& \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\& =\left|\begin{array}{ccc}-1 & 1 & -1 \\1 & -1 & 2 \\70 & -26 & -42\end{array}\right| \\\\& =-94+182-44=44\end{aligned}

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