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Mathematics · Vector Algebra

JEE Main 2025 — 3 April, Evening Shift — Question 44

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} and d⃗\vec{d} be a vector such that b⃗×d⃗=c⃗×d⃗\vec{b} \times \vec{d}=\vec{c} \times \vec{d}

and a⃗⋅d⃗=4\vec{a} \cdot \vec{d}=4. Then ∣(a⃗×d⃗)∣2|(\vec{a} \times \vec{d})|^{2} is equal to ____\_\_\_\_ -.

Answer: 128

Numerical answer — enter this value.

Step-by-step solution

b⃗×d⃗=c⃗×d⃗⇒(b⃗−c⃗)×d⃗=0\vec{b} \times \vec{d}=\vec{c} \times \vec{d} \Rightarrow(\vec{b}-\vec{c}) \times \vec{d}=0

⇒(b⃗−c⃗)\Rightarrow(\vec{b}-\vec{c}) is parallel to d⃗\vec{d}

⇒d⃗=λ(i^−2j^+k^)\Rightarrow \quad \vec{d}=\lambda(\hat{i}-2 \hat{j}+\hat{k})

∵a⃗⋅d⃗=4\because \vec{a} \cdot \vec{d}=4

⇒λ−4λ+λ=4\Rightarrow \lambda-4 \lambda+\lambda=4

⇒λ=−2⇒d⃗=−2i^+4j^−2k^\Rightarrow \lambda=-2 \Rightarrow \vec{d}=-2 \hat{i}+4 \hat{j}-2 \hat{k}

a⃗×d⃗=∣i^j^k^121−24−2∣=−8i^+0j^+8k^\vec{a} \times \vec{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & 1\\ -2 & 4 & -2\end{array}\right|=-8 \hat{i}+0 \hat{j}+8 \hat{k}

∣a⃗×d⃗∣2=128|\vec{a} \times \vec{d}|^{2}=128

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Mathematics
Chapter
Vector Algebra
Topic
Vector or Cross Product of Two Vectors