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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Vector Algebra

Let a = i +2 j + k , b =3 i -3 j +3 k , c =2 i - j +2 k and d be a...

Let a⃗=i^+2j^+k^,b⃗=3i^−3j^+3k^\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, c→=2i^−j^+2k^\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} and d→\overrightarrow{\mathrm{d}} be a vector such that b→×d→=c→×d→\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}} and a→⋅d→=4\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4. Then ∣(a→×d→)∣2|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}})|^2 is equal to ______ .

Correct answer

0

Step-by-step solution

b→×d→=c→×d→-and a→⋅d→=4⇒d→=λ(b→−c→)=λ(i^−2j^+k^)∵a→⋅d→=4⇒λ=−2 Also. ∣a→×d→∣2+∣a→⋅d→∣2=∣a→∣2∣d→∣2⇒∣a→×d→∣2=6×4×6−16=128\begin{aligned} & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}} \text{-and } \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4 \\ & \Rightarrow \overrightarrow{\mathrm{ d}}=\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4 \Rightarrow \lambda=-2 \\ & \text{ Also. }|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}|^2+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|^2=|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{ d}}|^2 \\ & \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}}|^2=6 \times 4 \times 6-16=128\end{aligned}
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