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Mathematics · Vector Algebra

JEE Main 2025 — 24 January, Evening Shift — Question 12

Let a→=3i^−j^+2k^,b→=a→×(i^−2k^)\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}-2 \hat{\mathrm{k}}) and c→=b→×k^\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}}. Then the projection of c⃗−2j^\vec{c}-2 \hat{j} on a⃗\vec{a} is:

  1. Option A:

    373 \sqrt{7}

  2. Option B:

    14\sqrt{14}

  3. Option C:

    2142 \sqrt{14}

    Correct
  4. Option D:

    272 \sqrt{7}

Answer: C

Step-by-step solution

b˙=a→×(i^−3k^)=∣i^j^k^3−1210−2∣=2i^+8j^+k^c→=b→×k^=8i^−2j^c→−2j^=8i^−4j^Projection   of  (c→−2j^)   on   a→=(c→−2j^)⋅a^=⟨8,−4,0⟩⋅⟨3,−1,2⟩14=2814=214\begin{aligned} \dot{\mathrm{b}} &= \overrightarrow{\mathrm{a}} \times (\hat{\mathrm{i}} - 3\hat{\mathrm{k}}) \\[6pt] &= \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\[3pt] 3 & -1 & 2 \\[3pt] 1 & 0 & -2 \end{vmatrix} \\[8pt] &= 2\hat{\mathrm{i}} + 8\hat{\mathrm{j}} + \hat{\mathrm{k}} \\[10pt] \overrightarrow{\mathrm{c}} &= \overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}} = 8\hat{\mathrm{i}} - 2\hat{\mathrm{j}} \\[8pt] \overrightarrow{\mathrm{c}} - 2\hat{\mathrm{j}} &= 8\hat{\mathrm{i}} - 4\hat{\mathrm{j}} \\[10pt] \text{Projection\; of\;} (\overrightarrow{\mathrm{c}} - 2\hat{\mathrm{j}}) \; \text{ on\; } \overrightarrow{\mathrm{a}} &= (\overrightarrow{\mathrm{c}} - 2\hat{\mathrm{j}}) \cdot \hat{\mathrm{a}} \\[6pt] &= \frac{\langle 8, -4, 0 \rangle \cdot \langle 3, -1, 2 \rangle}{\sqrt{14}} \\[8pt] &= \frac{28}{\sqrt{14}} = 2\sqrt{14} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Mathematics
Chapter
Vector Algebra
Topic
Projection & component of a vector along another vector.

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