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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Vector Algebra

Let a = i -2 j +3 k , b =2 i + j - k , c =λ i + j + k and v = a × b...

Let a⃗=i^−2j^+3k^,b⃗=2i^+j^−k^,c⃗=λi^+j^+k^\vec{a}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{b}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} and v⃗=a⃗×b⃗\vec{v}=\vec{a} \times \vec{b}. If v⃗⋅c⃗=11\vec{v} \cdot \vec{c}=11 and the length of the projection of b⃗\vec{b} on c⃗\vec{c} is pp, then 9p29 p^{2} is equal to

Options

  1. A12
  2. B6
  3. C4
  4. D9

Correct answer

A. 12

Step-by-step solution

v⃗=a⃗×b⃗=∣i^j^k^1−2321−1∣=−i^+7j^+5k^\vec{v} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = -\hat{i} + 7\hat{j} + 5\hat{k}.
v⃗⋅c⃗=−λ+7+5=11⇒λ=1\vec{v} \cdot \vec{c} = -\lambda + 7 + 5 = 11 \Rightarrow \lambda = 1.
c⃗=i^+j^+k^\vec{c} = \hat{i} + \hat{j} + \hat{k}, ∣c⃗∣=3|\vec{c}| = \sqrt{3}.
b⃗⋅c⃗=2+1−1=2\vec{b} \cdot \vec{c} = 2 + 1 - 1 = 2.
Projection p=∣b⃗⋅c⃗∣∣c⃗∣=23p = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} = \frac{2}{\sqrt{3}}.
9p2=9×43=129p^2 = 9 \times \frac{4}{3} = 12.
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