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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Vector Algebra

Let AB =2 i +4 j -5 k and AD = i +2 j +λ k , λ ∈ ℝ . Let the...

Let AB→=2i^+4j^−5k^\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k} and AD→=i^+2j^+λk^,λ∈R\overrightarrow{\mathrm{AD}}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \lambda \in \mathbb{R}. Let the projection of the vector v⃗=i^+j^+k^\vec{v}=\hat{i}+\hat{j}+\hat{k} on the diagonal AC→\overrightarrow{\mathrm{AC}} of the parallelogram ABCD be of length one unit. If α,β\alpha, \beta, where α>β\alpha>\beta, be the roots of the equation λ2x2−6λx+5=0\lambda^{2} x^{2}-6 \lambda x+5=0, then 2α−β2 \alpha-\beta is equal to

Options

  1. A3
  2. B4
  3. C1
  4. D6

Correct answer

A. 3

Step-by-step solution

AC⃗=AB⃗+AD⃗=3i^+6j^+(λ−5)k^\vec{AC} = \vec{AB} + \vec{AD} = 3\hat{i} + 6\hat{j} + (\lambda-5)\hat{k}.
Projection of v⃗=i^+j^+k^\vec{v} = \hat{i}+\hat{j}+\hat{k} on AC⃗\vec{AC}: ∣v⃗⋅AC⃗∣∣AC⃗∣=1\frac{|\vec{v} \cdot \vec{AC}|}{|\vec{AC}|} = 1.
v⃗⋅AC⃗=3+6+(λ−5)=λ+4\vec{v} \cdot \vec{AC} = 3+6+(\lambda-5) = \lambda+4, ∣AC⃗∣=45+(λ−5)2|\vec{AC}| = \sqrt{45+(\lambda-5)^2}.
(λ+4)2=45+(λ−5)2(\lambda+4)^2 = 45 + (\lambda-5)^2
⇒8λ+16=45−10λ+25\Rightarrow 8\lambda+16 = 45-10\lambda+25
⇒18λ=54⇒λ=3\Rightarrow 18\lambda = 54 \Rightarrow \lambda = 3.
Equation: 9x2−18x+5=09x^2 - 18x + 5 = 0
⇒(3x−1)(3x−5)=0\Rightarrow (3x-1)(3x-5) = 0
⇒x=13,53\Rightarrow x = \frac{1}{3}, \frac{5}{3}.
α=53, β=13\alpha = \frac{5}{3},\ \beta = \frac{1}{3}.
2α−β=103−13=32\alpha - \beta = \frac{10}{3} - \frac{1}{3} = 3.
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