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Vector Algebra question

2025 · 3 Apr · Shift 2 · Q49
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  5. /2025 · 3 Apr · Shift 2 · Q49

Vector Algebra question

2025 · 3 Apr · Shift 2 · Q49

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a⃗=i^+2j^+k^,b⃗=3i^−3j^+3k^,c⃗=2i^−j^+2k^\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}a=i^+2j^​+k^,b=3i^−3j^​+3k^,c=2i^−j^​+2k^ and d⃗\vec{d}d be a vector such that b⃗×d⃗=c⃗×d⃗\vec{b} \times \vec{d}=\vec{c} \times \vec{d}b×d=c×d and a⃗⋅d⃗=4\vec{a} \cdot \vec{d}=4a⋅d=4. Then ∣(a⃗×d⃗)∣2|(\vec{a} \times \vec{d})|^2∣(a×d)∣2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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